Prove that a group with $3$ elements is cyclic?
$begingroup$
Prove that a group with $3$ elements is cyclic.
I tried the case where $G={e,a,b}
$
and I kept trying multiplication
and finally I found that
$a^2$ must equal to $b$ and $b^2$ must equal to $a$.
Then $a^3=e$.
Are there any other methods
?
I have another question :
Prove that a group with $4$ elements may or may not be cyclic.
group-theory cyclic-groups
edited Jan 3 at 17:36
Thomas Shelby
3,7492525
asked Jan 3 at 17:16
El MoudenEl Mouden
11810
$endgroup$
|
show 1 more comment
$begingroup$
Prove that a group with $3$ elements is cyclic.
I tried the case where $G={e,a,b}
$
and I kept trying multiplication
and finally I found that
$a^2$ must equal to $b$ and $b^2$ must equal to $a$.
Then $a^3=e$.
Are there any other methods
?
I have another question :
Prove that a group with $4$ elements may or may not be cyclic.
group-theory cyclic-groups
edited Jan 3 at 17:36
Thomas Shelby
3,7492525
asked Jan 3 at 17:16
El MoudenEl Mouden
11810
$endgroup$
1
$begingroup$
Another method is to use the Lagrange's Theorem. Using Lagrange Theorem, you can prove that any group of prime order is cyclic.
$endgroup$
– Thomas Shelby
Jan 3 at 17:23
$begingroup$
I think that it doesn't take a lot of work to prove that any group of order $4$ (note that I omitted the word “finite”; are you aware of some infinite group of order $4$?) may by cyclic or not.
$endgroup$
– José Carlos Santos
Jan 3 at 17:31
$begingroup$
I recommend to read my answer here math.stackexchange.com/questions/1570641/…. A similar method works for a group of 4 elements.
$endgroup$
– Nicky Hekster
Jan 3 at 17:52
1
$begingroup$
@ThomasShelby: you mean Cauchy's theorem, right? Lagrange's theorem only gives $o(g)mid o(G)$, while Cauchy's theorem ensures $o(G)=pLongrightarrowexists g:o(g)=p$.
$endgroup$
– Jack D'Aurizio
Jan 3 at 20:27
1
$begingroup$
@ThomasShelby: all right, I get it. Since the identity is unique and $o(g)mid o(G)$, in a group with prime order all the elements except the identity are generators, fine.
$endgroup$
– Jack D'Aurizio
Jan 4 at 1:22
|
show 1 more comment
$begingroup$
Prove that a group with $3$ elements is cyclic.
I tried the case where $G={e,a,b}
$
and I kept trying multiplication
and finally I found that
$a^2$ must equal to $b$ and $b^2$ must equal to $a$.
Then $a^3=e$.
Are there any other methods
?
I have another question :
Prove that a group with $4$ elements may or may not be cyclic.
group-theory cyclic-groups
edited Jan 3 at 17:36
Thomas Shelby
3,7492525
asked Jan 3 at 17:16
El MoudenEl Mouden
11810
$endgroup$
Prove that a group with $3$ elements is cyclic.
I tried the case where $G={e,a,b}
$
and I kept trying multiplication
and finally I found that
$a^2$ must equal to $b$ and $b^2$ must equal to $a$.
Then $a^3=e$.
Are there any other methods
?
I have another question :
Prove that a group with $4$ elements may or may not be cyclic.
group-theory cyclic-groups
group-theory cyclic-groups
edited Jan 3 at 17:36
Thomas Shelby
3,7492525
asked Jan 3 at 17:16
El MoudenEl Mouden
11810
edited Jan 3 at 17:36
Thomas Shelby
3,7492525
asked Jan 3 at 17:16
El MoudenEl Mouden
11810
edited Jan 3 at 17:36
Thomas Shelby
3,7492525
edited Jan 3 at 17:36
Thomas Shelby
3,7492525
edited Jan 3 at 17:36
Thomas Shelby
3,7492525
3,7492525
asked Jan 3 at 17:16
El MoudenEl Mouden
11810
asked Jan 3 at 17:16
El MoudenEl Mouden
11810
asked Jan 3 at 17:16
El MoudenEl Mouden
11810
11810
1
$begingroup$
Another method is to use the Lagrange's Theorem. Using Lagrange Theorem, you can prove that any group of prime order is cyclic.
$endgroup$
– Thomas Shelby
Jan 3 at 17:23
$begingroup$
I think that it doesn't take a lot of work to prove that any group of order $4$ (note that I omitted the word “finite”; are you aware of some infinite group of order $4$?) may by cyclic or not.
$endgroup$
– José Carlos Santos
Jan 3 at 17:31
$begingroup$
I recommend to read my answer here math.stackexchange.com/questions/1570641/…. A similar method works for a group of 4 elements.
$endgroup$
– Nicky Hekster
Jan 3 at 17:52
1
$begingroup$
@ThomasShelby: you mean Cauchy's theorem, right? Lagrange's theorem only gives $o(g)mid o(G)$, while Cauchy's theorem ensures $o(G)=pLongrightarrowexists g:o(g)=p$.
$endgroup$
– Jack D'Aurizio
Jan 3 at 20:27
1
$begingroup$
@ThomasShelby: all right, I get it. Since the identity is unique and $o(g)mid o(G)$, in a group with prime order all the elements except the identity are generators, fine.
$endgroup$
– Jack D'Aurizio
Jan 4 at 1:22
|
show 1 more comment
1
$begingroup$
Another method is to use the Lagrange's Theorem. Using Lagrange Theorem, you can prove that any group of prime order is cyclic.
$endgroup$
– Thomas Shelby
Jan 3 at 17:23
$begingroup$
I think that it doesn't take a lot of work to prove that any group of order $4$ (note that I omitted the word “finite”; are you aware of some infinite group of order $4$?) may by cyclic or not.
$endgroup$
– José Carlos Santos
Jan 3 at 17:31
$begingroup$
I recommend to read my answer here math.stackexchange.com/questions/1570641/…. A similar method works for a group of 4 elements.
$endgroup$
– Nicky Hekster
Jan 3 at 17:52
1
$begingroup$
@ThomasShelby: you mean Cauchy's theorem, right? Lagrange's theorem only gives $o(g)mid o(G)$, while Cauchy's theorem ensures $o(G)=pLongrightarrowexists g:o(g)=p$.
$endgroup$
– Jack D'Aurizio
Jan 3 at 20:27
1
$begingroup$
@ThomasShelby: all right, I get it. Since the identity is unique and $o(g)mid o(G)$, in a group with prime order all the elements except the identity are generators, fine.
$endgroup$
– Jack D'Aurizio
Jan 4 at 1:22
1
1
$begingroup$
Another method is to use the Lagrange's Theorem. Using Lagrange Theorem, you can prove that any group of prime order is cyclic.
$endgroup$
– Thomas Shelby
Jan 3 at 17:23
$begingroup$
Another method is to use the Lagrange's Theorem. Using Lagrange Theorem, you can prove that any group of prime order is cyclic.
$endgroup$
– Thomas Shelby
Jan 3 at 17:23
$begingroup$
I think that it doesn't take a lot of work to prove that any group of order $4$ (note that I omitted the word “finite”; are you aware of some infinite group of order $4$?) may by cyclic or not.
$endgroup$
– José Carlos Santos
Jan 3 at 17:31
$begingroup$
I think that it doesn't take a lot of work to prove that any group of order $4$ (note that I omitted the word “finite”; are you aware of some infinite group of order $4$?) may by cyclic or not.
$endgroup$
– José Carlos Santos
Jan 3 at 17:31
$begingroup$
I recommend to read my answer here math.stackexchange.com/questions/1570641/…. A similar method works for a group of 4 elements.
$endgroup$
– Nicky Hekster
Jan 3 at 17:52
$begingroup$
I recommend to read my answer here math.stackexchange.com/questions/1570641/…. A similar method works for a group of 4 elements.
$endgroup$
– Nicky Hekster
Jan 3 at 17:52
1
1
$begingroup$
@ThomasShelby: you mean Cauchy's theorem, right? Lagrange's theorem only gives $o(g)mid o(G)$, while Cauchy's theorem ensures $o(G)=pLongrightarrowexists g:o(g)=p$.
$endgroup$
– Jack D'Aurizio
Jan 3 at 20:27
$begingroup$
@ThomasShelby: you mean Cauchy's theorem, right? Lagrange's theorem only gives $o(g)mid o(G)$, while Cauchy's theorem ensures $o(G)=pLongrightarrowexists g:o(g)=p$.
$endgroup$
– Jack D'Aurizio
Jan 3 at 20:27
1
1
$begingroup$
@ThomasShelby: all right, I get it. Since the identity is unique and $o(g)mid o(G)$, in a group with prime order all the elements except the identity are generators, fine.
$endgroup$
– Jack D'Aurizio
Jan 4 at 1:22
$begingroup$
@ThomasShelby: all right, I get it. Since the identity is unique and $o(g)mid o(G)$, in a group with prime order all the elements except the identity are generators, fine.
$endgroup$
– Jack D'Aurizio
Jan 4 at 1:22
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
According to Lagrange's theorem, a group with 3 elements can have only the trivial subgroups. Because 3 is prime only 1 and 3 divide it. So, only possible subgroups are ${e}$ with one element and $G$ with three. For $x in G$ that is not $e$ a group generated with this element,
$<x>$ must be whole $G$. So $G$ is cyclic.
A group with 4 elements can have untrivial subgroups as 4 isn't prime. For instance $mathbb{Z_4} = {0, 1, 2, 3}$ with $+$ is cyclic. It is generated with the element $1$. But group $mathbb{Z_2} times mathbb{Z_2} = {(0, 0), (1, 0), (0, 1), (1, 1)}$ with $+$ defined as $(a, b) + (a', b') = (a + a', b + b')$ is not cyclic.
answered Jan 3 at 18:13
CoupeauCoupeau
1296
$endgroup$
$begingroup$
can you explaine in more terms " For x∈G that is not e a group generated with this element, <x> must be whole G. So G is cyclic."
$endgroup$
– El Mouden
Jan 8 at 12:23
$begingroup$
We can take any element (that is not the unit $e$) $x in G$ and by multiplying it with itself, we will get every other element of this group. So let $x in G$. $x x$ is another element that must be in the group but is different from $x$ since we said $x$ is not a unit. So, now we have three distinct elements: $e, x, x^2$. We also know, that since our group can only have either one or three elements, $x^2$ can not be the unit. Now we have the whole $G$.
$endgroup$
– Coupeau
Jan 9 at 17:49
add a comment |
$begingroup$
Instead of “keeping multiplying” it's easier to fill the Cayley diagram: in every row and column every element must appear.
begin{array}{c|ccc}
& e & a & b \ hline
e & e & a & b \
a & a & \
b & b & \
end{array}
In the slot corresponding to $a^2$ you cannot put $e$, because otherwise the slot for $ab$ would contain $b$. Hence $a^2=b$
begin{array}{c|ccc}
& e & a & b \ hline
e & e & a & b \
a & a & b\
b & b & \
end{array}
and now the diagram has a unique completion
begin{array}{c|ccc}
& e & a & b \ hline
e & e & a & b \
a & a & b & e \
b & b & e & a\
end{array}
Then $a^3=a^2a=ba=e$.
You can try your hand with a four element group and see that the diagram admits different completions.
begin{array}{c|cccc}
& e & a & b & c \ hline
e & e & a & b & c \
a & a & \
b & b & \
c & c &
end{array}
In the slot for $a^2$ we can put any of $e$, $b$ or $c$. Let's try with $e$:
begin{array}{c|cccc}
& e & a & b & c \ hline
e & e & a & b & c \
a & a & e\
b & b & \
c & c &
end{array}
Then we are forced to put in the following
begin{array}{c|cccc}
& e & a & b & c \ hline
e & e & a & b & c \
a & a & e & c & b \
b & b & c\
c & c & b
end{array}
For $b^2$ we can have either $e$ or $c$. Let's try $e$:
begin{array}{c|cccc}
& e & a & b & c \ hline
e & e & a & b & c \
a & a & e & c & b \
b & b & c & e\
c & c & b &
end{array}
Now we can complete:
begin{array}{c|cccc}
& e & a & b & c \ hline
e & e & a & b & c \
a & a & e & c & b \
b & b & c & e & a\
c & c & b & a & e
end{array}
It would be very tedious to verify that this diagram indeed produces a group. It does and is the product of two cyclic groups of order two; it is also the subgroup of $S_4$ given by ${id,(12)(34),(13)(24),(14)(23)}$.
answered Jan 3 at 18:26
egregegreg
183k1486205
$endgroup$
$begingroup$
I need another proof, without using the Cayley diagram
$endgroup$
– El Mouden
Jan 8 at 12:25
$begingroup$
@ElMouden Then using Lagrange's theorem is what you need.
$endgroup$
– egreg
Jan 8 at 12:59
$begingroup$
Oh Thanks Finally i did it
$endgroup$
– El Mouden
Jan 9 at 14:07
add a comment |
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2 Answers
2
active
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2 Answers
2
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active
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active
oldest
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$begingroup$
According to Lagrange's theorem, a group with 3 elements can have only the trivial subgroups. Because 3 is prime only 1 and 3 divide it. So, only possible subgroups are ${e}$ with one element and $G$ with three. For $x in G$ that is not $e$ a group generated with this element,
$<x>$ must be whole $G$. So $G$ is cyclic.
A group with 4 elements can have untrivial subgroups as 4 isn't prime. For instance $mathbb{Z_4} = {0, 1, 2, 3}$ with $+$ is cyclic. It is generated with the element $1$. But group $mathbb{Z_2} times mathbb{Z_2} = {(0, 0), (1, 0), (0, 1), (1, 1)}$ with $+$ defined as $(a, b) + (a', b') = (a + a', b + b')$ is not cyclic.
answered Jan 3 at 18:13
CoupeauCoupeau
1296
$endgroup$
$begingroup$
can you explaine in more terms " For x∈G that is not e a group generated with this element, <x> must be whole G. So G is cyclic."
$endgroup$
– El Mouden
Jan 8 at 12:23
$begingroup$
We can take any element (that is not the unit $e$) $x in G$ and by multiplying it with itself, we will get every other element of this group. So let $x in G$. $x x$ is another element that must be in the group but is different from $x$ since we said $x$ is not a unit. So, now we have three distinct elements: $e, x, x^2$. We also know, that since our group can only have either one or three elements, $x^2$ can not be the unit. Now we have the whole $G$.
$endgroup$
– Coupeau
Jan 9 at 17:49
add a comment |
$begingroup$
According to Lagrange's theorem, a group with 3 elements can have only the trivial subgroups. Because 3 is prime only 1 and 3 divide it. So, only possible subgroups are ${e}$ with one element and $G$ with three. For $x in G$ that is not $e$ a group generated with this element,
$<x>$ must be whole $G$. So $G$ is cyclic.
A group with 4 elements can have untrivial subgroups as 4 isn't prime. For instance $mathbb{Z_4} = {0, 1, 2, 3}$ with $+$ is cyclic. It is generated with the element $1$. But group $mathbb{Z_2} times mathbb{Z_2} = {(0, 0), (1, 0), (0, 1), (1, 1)}$ with $+$ defined as $(a, b) + (a', b') = (a + a', b + b')$ is not cyclic.
answered Jan 3 at 18:13
CoupeauCoupeau
1296
$endgroup$
$begingroup$
can you explaine in more terms " For x∈G that is not e a group generated with this element, <x> must be whole G. So G is cyclic."
$endgroup$
– El Mouden
Jan 8 at 12:23
$begingroup$
We can take any element (that is not the unit $e$) $x in G$ and by multiplying it with itself, we will get every other element of this group. So let $x in G$. $x x$ is another element that must be in the group but is different from $x$ since we said $x$ is not a unit. So, now we have three distinct elements: $e, x, x^2$. We also know, that since our group can only have either one or three elements, $x^2$ can not be the unit. Now we have the whole $G$.
$endgroup$
– Coupeau
Jan 9 at 17:49
add a comment |
$begingroup$
According to Lagrange's theorem, a group with 3 elements can have only the trivial subgroups. Because 3 is prime only 1 and 3 divide it. So, only possible subgroups are ${e}$ with one element and $G$ with three. For $x in G$ that is not $e$ a group generated with this element,
$<x>$ must be whole $G$. So $G$ is cyclic.
A group with 4 elements can have untrivial subgroups as 4 isn't prime. For instance $mathbb{Z_4} = {0, 1, 2, 3}$ with $+$ is cyclic. It is generated with the element $1$. But group $mathbb{Z_2} times mathbb{Z_2} = {(0, 0), (1, 0), (0, 1), (1, 1)}$ with $+$ defined as $(a, b) + (a', b') = (a + a', b + b')$ is not cyclic.
answered Jan 3 at 18:13
CoupeauCoupeau
1296
$endgroup$
According to Lagrange's theorem, a group with 3 elements can have only the trivial subgroups. Because 3 is prime only 1 and 3 divide it. So, only possible subgroups are ${e}$ with one element and $G$ with three. For $x in G$ that is not $e$ a group generated with this element,
$<x>$ must be whole $G$. So $G$ is cyclic.
A group with 4 elements can have untrivial subgroups as 4 isn't prime. For instance $mathbb{Z_4} = {0, 1, 2, 3}$ with $+$ is cyclic. It is generated with the element $1$. But group $mathbb{Z_2} times mathbb{Z_2} = {(0, 0), (1, 0), (0, 1), (1, 1)}$ with $+$ defined as $(a, b) + (a', b') = (a + a', b + b')$ is not cyclic.
answered Jan 3 at 18:13
CoupeauCoupeau
1296
answered Jan 3 at 18:13
CoupeauCoupeau
1296
answered Jan 3 at 18:13
CoupeauCoupeau
1296
answered Jan 3 at 18:13
CoupeauCoupeau
1296
1296
$begingroup$
can you explaine in more terms " For x∈G that is not e a group generated with this element, <x> must be whole G. So G is cyclic."
$endgroup$
– El Mouden
Jan 8 at 12:23
$begingroup$
We can take any element (that is not the unit $e$) $x in G$ and by multiplying it with itself, we will get every other element of this group. So let $x in G$. $x x$ is another element that must be in the group but is different from $x$ since we said $x$ is not a unit. So, now we have three distinct elements: $e, x, x^2$. We also know, that since our group can only have either one or three elements, $x^2$ can not be the unit. Now we have the whole $G$.
$endgroup$
– Coupeau
Jan 9 at 17:49
add a comment |
$begingroup$
can you explaine in more terms " For x∈G that is not e a group generated with this element, <x> must be whole G. So G is cyclic."
$endgroup$
– El Mouden
Jan 8 at 12:23
$begingroup$
We can take any element (that is not the unit $e$) $x in G$ and by multiplying it with itself, we will get every other element of this group. So let $x in G$. $x x$ is another element that must be in the group but is different from $x$ since we said $x$ is not a unit. So, now we have three distinct elements: $e, x, x^2$. We also know, that since our group can only have either one or three elements, $x^2$ can not be the unit. Now we have the whole $G$.
$endgroup$
– Coupeau
Jan 9 at 17:49
$begingroup$
can you explaine in more terms " For x∈G that is not e a group generated with this element, <x> must be whole G. So G is cyclic."
$endgroup$
– El Mouden
Jan 8 at 12:23
$begingroup$
can you explaine in more terms " For x∈G that is not e a group generated with this element, <x> must be whole G. So G is cyclic."
$endgroup$
– El Mouden
Jan 8 at 12:23
$begingroup$
We can take any element (that is not the unit $e$) $x in G$ and by multiplying it with itself, we will get every other element of this group. So let $x in G$. $x x$ is another element that must be in the group but is different from $x$ since we said $x$ is not a unit. So, now we have three distinct elements: $e, x, x^2$. We also know, that since our group can only have either one or three elements, $x^2$ can not be the unit. Now we have the whole $G$.
$endgroup$
– Coupeau
Jan 9 at 17:49
$begingroup$
We can take any element (that is not the unit $e$) $x in G$ and by multiplying it with itself, we will get every other element of this group. So let $x in G$. $x x$ is another element that must be in the group but is different from $x$ since we said $x$ is not a unit. So, now we have three distinct elements: $e, x, x^2$. We also know, that since our group can only have either one or three elements, $x^2$ can not be the unit. Now we have the whole $G$.
$endgroup$
– Coupeau
Jan 9 at 17:49
add a comment |
$begingroup$
Instead of “keeping multiplying” it's easier to fill the Cayley diagram: in every row and column every element must appear.
begin{array}{c|ccc}
& e & a & b \ hline
e & e & a & b \
a & a & \
b & b & \
end{array}
In the slot corresponding to $a^2$ you cannot put $e$, because otherwise the slot for $ab$ would contain $b$. Hence $a^2=b$
begin{array}{c|ccc}
& e & a & b \ hline
e & e & a & b \
a & a & b\
b & b & \
end{array}
and now the diagram has a unique completion
begin{array}{c|ccc}
& e & a & b \ hline
e & e & a & b \
a & a & b & e \
b & b & e & a\
end{array}
Then $a^3=a^2a=ba=e$.
You can try your hand with a four element group and see that the diagram admits different completions.
begin{array}{c|cccc}
& e & a & b & c \ hline
e & e & a & b & c \
a & a & \
b & b & \
c & c &
end{array}
In the slot for $a^2$ we can put any of $e$, $b$ or $c$. Let's try with $e$:
begin{array}{c|cccc}
& e & a & b & c \ hline
e & e & a & b & c \
a & a & e\
b & b & \
c & c &
end{array}
Then we are forced to put in the following
begin{array}{c|cccc}
& e & a & b & c \ hline
e & e & a & b & c \
a & a & e & c & b \
b & b & c\
c & c & b
end{array}
For $b^2$ we can have either $e$ or $c$. Let's try $e$:
begin{array}{c|cccc}
& e & a & b & c \ hline
e & e & a & b & c \
a & a & e & c & b \
b & b & c & e\
c & c & b &
end{array}
Now we can complete:
begin{array}{c|cccc}
& e & a & b & c \ hline
e & e & a & b & c \
a & a & e & c & b \
b & b & c & e & a\
c & c & b & a & e
end{array}
It would be very tedious to verify that this diagram indeed produces a group. It does and is the product of two cyclic groups of order two; it is also the subgroup of $S_4$ given by ${id,(12)(34),(13)(24),(14)(23)}$.
answered Jan 3 at 18:26
egregegreg
183k1486205
$endgroup$
$begingroup$
I need another proof, without using the Cayley diagram
$endgroup$
– El Mouden
Jan 8 at 12:25
$begingroup$
@ElMouden Then using Lagrange's theorem is what you need.
$endgroup$
– egreg
Jan 8 at 12:59
$begingroup$
Oh Thanks Finally i did it
$endgroup$
– El Mouden
Jan 9 at 14:07
add a comment |
$begingroup$
Instead of “keeping multiplying” it's easier to fill the Cayley diagram: in every row and column every element must appear.
begin{array}{c|ccc}
& e & a & b \ hline
e & e & a & b \
a & a & \
b & b & \
end{array}
In the slot corresponding to $a^2$ you cannot put $e$, because otherwise the slot for $ab$ would contain $b$. Hence $a^2=b$
begin{array}{c|ccc}
& e & a & b \ hline
e & e & a & b \
a & a & b\
b & b & \
end{array}
and now the diagram has a unique completion
begin{array}{c|ccc}
& e & a & b \ hline
e & e & a & b \
a & a & b & e \
b & b & e & a\
end{array}
Then $a^3=a^2a=ba=e$.
You can try your hand with a four element group and see that the diagram admits different completions.
begin{array}{c|cccc}
& e & a & b & c \ hline
e & e & a & b & c \
a & a & \
b & b & \
c & c &
end{array}
In the slot for $a^2$ we can put any of $e$, $b$ or $c$. Let's try with $e$:
begin{array}{c|cccc}
& e & a & b & c \ hline
e & e & a & b & c \
a & a & e\
b & b & \
c & c &
end{array}
Then we are forced to put in the following
begin{array}{c|cccc}
& e & a & b & c \ hline
e & e & a & b & c \
a & a & e & c & b \
b & b & c\
c & c & b
end{array}
For $b^2$ we can have either $e$ or $c$. Let's try $e$:
begin{array}{c|cccc}
& e & a & b & c \ hline
e & e & a & b & c \
a & a & e & c & b \
b & b & c & e\
c & c & b &
end{array}
Now we can complete:
begin{array}{c|cccc}
& e & a & b & c \ hline
e & e & a & b & c \
a & a & e & c & b \
b & b & c & e & a\
c & c & b & a & e
end{array}
It would be very tedious to verify that this diagram indeed produces a group. It does and is the product of two cyclic groups of order two; it is also the subgroup of $S_4$ given by ${id,(12)(34),(13)(24),(14)(23)}$.
answered Jan 3 at 18:26
egregegreg
183k1486205
$endgroup$
$begingroup$
I need another proof, without using the Cayley diagram
$endgroup$
– El Mouden
Jan 8 at 12:25
$begingroup$
@ElMouden Then using Lagrange's theorem is what you need.
$endgroup$
– egreg
Jan 8 at 12:59
$begingroup$
Oh Thanks Finally i did it
$endgroup$
– El Mouden
Jan 9 at 14:07
add a comment |
$begingroup$
Instead of “keeping multiplying” it's easier to fill the Cayley diagram: in every row and column every element must appear.
begin{array}{c|ccc}
& e & a & b \ hline
e & e & a & b \
a & a & \
b & b & \
end{array}
In the slot corresponding to $a^2$ you cannot put $e$, because otherwise the slot for $ab$ would contain $b$. Hence $a^2=b$
begin{array}{c|ccc}
& e & a & b \ hline
e & e & a & b \
a & a & b\
b & b & \
end{array}
and now the diagram has a unique completion
begin{array}{c|ccc}
& e & a & b \ hline
e & e & a & b \
a & a & b & e \
b & b & e & a\
end{array}
Then $a^3=a^2a=ba=e$.
You can try your hand with a four element group and see that the diagram admits different completions.
begin{array}{c|cccc}
& e & a & b & c \ hline
e & e & a & b & c \
a & a & \
b & b & \
c & c &
end{array}
In the slot for $a^2$ we can put any of $e$, $b$ or $c$. Let's try with $e$:
begin{array}{c|cccc}
& e & a & b & c \ hline
e & e & a & b & c \
a & a & e\
b & b & \
c & c &
end{array}
Then we are forced to put in the following
begin{array}{c|cccc}
& e & a & b & c \ hline
e & e & a & b & c \
a & a & e & c & b \
b & b & c\
c & c & b
end{array}
For $b^2$ we can have either $e$ or $c$. Let's try $e$:
begin{array}{c|cccc}
& e & a & b & c \ hline
e & e & a & b & c \
a & a & e & c & b \
b & b & c & e\
c & c & b &
end{array}
Now we can complete:
begin{array}{c|cccc}
& e & a & b & c \ hline
e & e & a & b & c \
a & a & e & c & b \
b & b & c & e & a\
c & c & b & a & e
end{array}
It would be very tedious to verify that this diagram indeed produces a group. It does and is the product of two cyclic groups of order two; it is also the subgroup of $S_4$ given by ${id,(12)(34),(13)(24),(14)(23)}$.
answered Jan 3 at 18:26
egregegreg
183k1486205
$endgroup$
Instead of “keeping multiplying” it's easier to fill the Cayley diagram: in every row and column every element must appear.
begin{array}{c|ccc}
& e & a & b \ hline
e & e & a & b \
a & a & \
b & b & \
end{array}
In the slot corresponding to $a^2$ you cannot put $e$, because otherwise the slot for $ab$ would contain $b$. Hence $a^2=b$
begin{array}{c|ccc}
& e & a & b \ hline
e & e & a & b \
a & a & b\
b & b & \
end{array}
and now the diagram has a unique completion
begin{array}{c|ccc}
& e & a & b \ hline
e & e & a & b \
a & a & b & e \
b & b & e & a\
end{array}
Then $a^3=a^2a=ba=e$.
You can try your hand with a four element group and see that the diagram admits different completions.
begin{array}{c|cccc}
& e & a & b & c \ hline
e & e & a & b & c \
a & a & \
b & b & \
c & c &
end{array}
In the slot for $a^2$ we can put any of $e$, $b$ or $c$. Let's try with $e$:
begin{array}{c|cccc}
& e & a & b & c \ hline
e & e & a & b & c \
a & a & e\
b & b & \
c & c &
end{array}
Then we are forced to put in the following
begin{array}{c|cccc}
& e & a & b & c \ hline
e & e & a & b & c \
a & a & e & c & b \
b & b & c\
c & c & b
end{array}
For $b^2$ we can have either $e$ or $c$. Let's try $e$:
begin{array}{c|cccc}
& e & a & b & c \ hline
e & e & a & b & c \
a & a & e & c & b \
b & b & c & e\
c & c & b &
end{array}
Now we can complete:
begin{array}{c|cccc}
& e & a & b & c \ hline
e & e & a & b & c \
a & a & e & c & b \
b & b & c & e & a\
c & c & b & a & e
end{array}
It would be very tedious to verify that this diagram indeed produces a group. It does and is the product of two cyclic groups of order two; it is also the subgroup of $S_4$ given by ${id,(12)(34),(13)(24),(14)(23)}$.
answered Jan 3 at 18:26
egregegreg
183k1486205
edited Jan 3 at 18:33
answered Jan 3 at 18:26
egregegreg
183k1486205
answered Jan 3 at 18:26
egregegreg
183k1486205
answered Jan 3 at 18:26
egregegreg
183k1486205
183k1486205
$begingroup$
I need another proof, without using the Cayley diagram
$endgroup$
– El Mouden
Jan 8 at 12:25
$begingroup$
@ElMouden Then using Lagrange's theorem is what you need.
$endgroup$
– egreg
Jan 8 at 12:59
$begingroup$
Oh Thanks Finally i did it
$endgroup$
– El Mouden
Jan 9 at 14:07
add a comment |
$begingroup$
I need another proof, without using the Cayley diagram
$endgroup$
– El Mouden
Jan 8 at 12:25
$begingroup$
@ElMouden Then using Lagrange's theorem is what you need.
$endgroup$
– egreg
Jan 8 at 12:59
$begingroup$
Oh Thanks Finally i did it
$endgroup$
– El Mouden
Jan 9 at 14:07
$begingroup$
I need another proof, without using the Cayley diagram
$endgroup$
– El Mouden
Jan 8 at 12:25
$begingroup$
I need another proof, without using the Cayley diagram
$endgroup$
– El Mouden
Jan 8 at 12:25
$begingroup$
@ElMouden Then using Lagrange's theorem is what you need.
$endgroup$
– egreg
Jan 8 at 12:59
$begingroup$
@ElMouden Then using Lagrange's theorem is what you need.
$endgroup$
– egreg
Jan 8 at 12:59
$begingroup$
Oh Thanks Finally i did it
$endgroup$
– El Mouden
Jan 9 at 14:07
$begingroup$
Oh Thanks Finally i did it
$endgroup$
– El Mouden
Jan 9 at 14:07
add a comment |
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1
$begingroup$
Another method is to use the Lagrange's Theorem. Using Lagrange Theorem, you can prove that any group of prime order is cyclic.
$endgroup$
– Thomas Shelby
Jan 3 at 17:23
$begingroup$
I think that it doesn't take a lot of work to prove that any group of order $4$ (note that I omitted the word “finite”; are you aware of some infinite group of order $4$?) may by cyclic or not.
$endgroup$
– José Carlos Santos
Jan 3 at 17:31
$begingroup$
I recommend to read my answer here math.stackexchange.com/questions/1570641/…. A similar method works for a group of 4 elements.
$endgroup$
– Nicky Hekster
Jan 3 at 17:52
1
$begingroup$
@ThomasShelby: you mean Cauchy's theorem, right? Lagrange's theorem only gives $o(g)mid o(G)$, while Cauchy's theorem ensures $o(G)=pLongrightarrowexists g:o(g)=p$.
$endgroup$
– Jack D'Aurizio
Jan 3 at 20:27
1
$begingroup$
@ThomasShelby: all right, I get it. Since the identity is unique and $o(g)mid o(G)$, in a group with prime order all the elements except the identity are generators, fine.
$endgroup$
– Jack D'Aurizio
Jan 4 at 1:22