Subgroups of $G^n$, where $G$ is a $p$-group
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Let $G$ be a finite $p$-group and let $n > 0$. Let $G^n$ be the direct product of $n$ copies of $G$.
Are all subgroups of $G^n$ isomorphic to $H_1 times dotsm times H_n$ for some subgroups $H_1, dots, H_n$
of $G$?
Comments. The question is being isomorphic to a direct product of subgroups, and not being equal to a direct product of subgroups. A negative answer to a similar question for $G = S_3$ and $n = 2$ was given here.
This question and its answer, which relies on Goursat's lemma might also be relevant.
P.S. I am especially interested in the case $p = 2$.
group-theory finite-groups p-groups
edited Nov 22 at 12:48
Derek Holt
52.3k53570
asked Nov 22 at 12:12
J.-E. Pin
18.3k21754
|
show 2 more comments
up vote
3
down vote
favorite
Let $G$ be a finite $p$-group and let $n > 0$. Let $G^n$ be the direct product of $n$ copies of $G$.
Are all subgroups of $G^n$ isomorphic to $H_1 times dotsm times H_n$ for some subgroups $H_1, dots, H_n$
of $G$?
Comments. The question is being isomorphic to a direct product of subgroups, and not being equal to a direct product of subgroups. A negative answer to a similar question for $G = S_3$ and $n = 2$ was given here.
This question and its answer, which relies on Goursat's lemma might also be relevant.
P.S. I am especially interested in the case $p = 2$.
group-theory finite-groups p-groups
edited Nov 22 at 12:48
Derek Holt
52.3k53570
asked Nov 22 at 12:12
J.-E. Pin
18.3k21754
Have you run any computer checks for small values of $|G|$ and $n$?
– the_fox
Nov 22 at 12:25
I tried a few examples on GAP without finding any counterexample so far.
– J.-E. Pin
Nov 22 at 12:34
I doubt this is true, but I could be wrong. Include the code you have used to check this, if you want, and we can split the workload. (I was just about to start thinking how to write a piece of GAP code that would work, so it would save me some time.)
– the_fox
Nov 22 at 12:41
I have replaced "Is any" in your question by "Are all". The word "any" is very often ambiguous.
– Derek Holt
Nov 22 at 12:49
1
I think $D_8 times D_8$ has a subdirect product of index 2 that does not decompose as a nontrivial direct product. (Here $D_8$ means dihedral group of order $8$.)
– Derek Holt
Nov 22 at 12:55
|
show 2 more comments
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let $G$ be a finite $p$-group and let $n > 0$. Let $G^n$ be the direct product of $n$ copies of $G$.
Are all subgroups of $G^n$ isomorphic to $H_1 times dotsm times H_n$ for some subgroups $H_1, dots, H_n$
of $G$?
Comments. The question is being isomorphic to a direct product of subgroups, and not being equal to a direct product of subgroups. A negative answer to a similar question for $G = S_3$ and $n = 2$ was given here.
This question and its answer, which relies on Goursat's lemma might also be relevant.
P.S. I am especially interested in the case $p = 2$.
group-theory finite-groups p-groups
edited Nov 22 at 12:48
Derek Holt
52.3k53570
asked Nov 22 at 12:12
J.-E. Pin
18.3k21754
Let $G$ be a finite $p$-group and let $n > 0$. Let $G^n$ be the direct product of $n$ copies of $G$.
Are all subgroups of $G^n$ isomorphic to $H_1 times dotsm times H_n$ for some subgroups $H_1, dots, H_n$
of $G$?
Comments. The question is being isomorphic to a direct product of subgroups, and not being equal to a direct product of subgroups. A negative answer to a similar question for $G = S_3$ and $n = 2$ was given here.
This question and its answer, which relies on Goursat's lemma might also be relevant.
P.S. I am especially interested in the case $p = 2$.
group-theory finite-groups p-groups
group-theory finite-groups p-groups
edited Nov 22 at 12:48
Derek Holt
52.3k53570
asked Nov 22 at 12:12
J.-E. Pin
18.3k21754
edited Nov 22 at 12:48
Derek Holt
52.3k53570
asked Nov 22 at 12:12
J.-E. Pin
18.3k21754
edited Nov 22 at 12:48
Derek Holt
52.3k53570
edited Nov 22 at 12:48
Derek Holt
52.3k53570
edited Nov 22 at 12:48
Derek Holt
52.3k53570
52.3k53570
asked Nov 22 at 12:12
J.-E. Pin
18.3k21754
asked Nov 22 at 12:12
J.-E. Pin
18.3k21754
asked Nov 22 at 12:12
J.-E. Pin
18.3k21754
18.3k21754
Have you run any computer checks for small values of $|G|$ and $n$?
– the_fox
Nov 22 at 12:25
I tried a few examples on GAP without finding any counterexample so far.
– J.-E. Pin
Nov 22 at 12:34
I doubt this is true, but I could be wrong. Include the code you have used to check this, if you want, and we can split the workload. (I was just about to start thinking how to write a piece of GAP code that would work, so it would save me some time.)
– the_fox
Nov 22 at 12:41
I have replaced "Is any" in your question by "Are all". The word "any" is very often ambiguous.
– Derek Holt
Nov 22 at 12:49
1
I think $D_8 times D_8$ has a subdirect product of index 2 that does not decompose as a nontrivial direct product. (Here $D_8$ means dihedral group of order $8$.)
– Derek Holt
Nov 22 at 12:55
|
show 2 more comments
Have you run any computer checks for small values of $|G|$ and $n$?
– the_fox
Nov 22 at 12:25
I tried a few examples on GAP without finding any counterexample so far.
– J.-E. Pin
Nov 22 at 12:34
I doubt this is true, but I could be wrong. Include the code you have used to check this, if you want, and we can split the workload. (I was just about to start thinking how to write a piece of GAP code that would work, so it would save me some time.)
– the_fox
Nov 22 at 12:41
I have replaced "Is any" in your question by "Are all". The word "any" is very often ambiguous.
– Derek Holt
Nov 22 at 12:49
1
I think $D_8 times D_8$ has a subdirect product of index 2 that does not decompose as a nontrivial direct product. (Here $D_8$ means dihedral group of order $8$.)
– Derek Holt
Nov 22 at 12:55
Have you run any computer checks for small values of $|G|$ and $n$?
– the_fox
Nov 22 at 12:25
Have you run any computer checks for small values of $|G|$ and $n$?
– the_fox
Nov 22 at 12:25
I tried a few examples on GAP without finding any counterexample so far.
– J.-E. Pin
Nov 22 at 12:34
I tried a few examples on GAP without finding any counterexample so far.
– J.-E. Pin
Nov 22 at 12:34
I doubt this is true, but I could be wrong. Include the code you have used to check this, if you want, and we can split the workload. (I was just about to start thinking how to write a piece of GAP code that would work, so it would save me some time.)
– the_fox
Nov 22 at 12:41
I doubt this is true, but I could be wrong. Include the code you have used to check this, if you want, and we can split the workload. (I was just about to start thinking how to write a piece of GAP code that would work, so it would save me some time.)
– the_fox
Nov 22 at 12:41
I have replaced "Is any" in your question by "Are all". The word "any" is very often ambiguous.
– Derek Holt
Nov 22 at 12:49
I have replaced "Is any" in your question by "Are all". The word "any" is very often ambiguous.
– Derek Holt
Nov 22 at 12:49
1
1
I think $D_8 times D_8$ has a subdirect product of index 2 that does not decompose as a nontrivial direct product. (Here $D_8$ means dihedral group of order $8$.)
– Derek Holt
Nov 22 at 12:55
I think $D_8 times D_8$ has a subdirect product of index 2 that does not decompose as a nontrivial direct product. (Here $D_8$ means dihedral group of order $8$.)
– Derek Holt
Nov 22 at 12:55
|
show 2 more comments
1 Answer
1
active
oldest
votes
up vote
5
down vote
accepted
Let $G=D_8 = langle a,b mid a^4=b^2=(ab)^2=1 rangle$, $G^2 = langle a_1,b_1rangle times langle a_2,b_2 rangle$ and $H = langle a_1a_2,b_1,b_2,a_1^2,a_2^2 rangle le G$ with $|H|=32$.
If $H cong H_1 times H_2$, with $H_1$ and $H_2$ isomorphic to subgroups of $D_8$, then we must have $|H_1|=8$, $|H_2|=4$ (or vice versa), so $H_2$ is abelian, and hence $|Z(H_1 times H_2)| =8$. But you can check that $Z(H)=Z(G) = langle a_1^2,a_2^2 rangle$ has order $4$.
In fact $H$ is indecomposable.
answered Nov 22 at 14:10
Derek Holt
52.3k53570
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
Let $G=D_8 = langle a,b mid a^4=b^2=(ab)^2=1 rangle$, $G^2 = langle a_1,b_1rangle times langle a_2,b_2 rangle$ and $H = langle a_1a_2,b_1,b_2,a_1^2,a_2^2 rangle le G$ with $|H|=32$.
If $H cong H_1 times H_2$, with $H_1$ and $H_2$ isomorphic to subgroups of $D_8$, then we must have $|H_1|=8$, $|H_2|=4$ (or vice versa), so $H_2$ is abelian, and hence $|Z(H_1 times H_2)| =8$. But you can check that $Z(H)=Z(G) = langle a_1^2,a_2^2 rangle$ has order $4$.
In fact $H$ is indecomposable.
answered Nov 22 at 14:10
Derek Holt
52.3k53570
add a comment |
up vote
5
down vote
accepted
Let $G=D_8 = langle a,b mid a^4=b^2=(ab)^2=1 rangle$, $G^2 = langle a_1,b_1rangle times langle a_2,b_2 rangle$ and $H = langle a_1a_2,b_1,b_2,a_1^2,a_2^2 rangle le G$ with $|H|=32$.
If $H cong H_1 times H_2$, with $H_1$ and $H_2$ isomorphic to subgroups of $D_8$, then we must have $|H_1|=8$, $|H_2|=4$ (or vice versa), so $H_2$ is abelian, and hence $|Z(H_1 times H_2)| =8$. But you can check that $Z(H)=Z(G) = langle a_1^2,a_2^2 rangle$ has order $4$.
In fact $H$ is indecomposable.
answered Nov 22 at 14:10
Derek Holt
52.3k53570
add a comment |
up vote
5
down vote
accepted
up vote
5
down vote
accepted
Let $G=D_8 = langle a,b mid a^4=b^2=(ab)^2=1 rangle$, $G^2 = langle a_1,b_1rangle times langle a_2,b_2 rangle$ and $H = langle a_1a_2,b_1,b_2,a_1^2,a_2^2 rangle le G$ with $|H|=32$.
If $H cong H_1 times H_2$, with $H_1$ and $H_2$ isomorphic to subgroups of $D_8$, then we must have $|H_1|=8$, $|H_2|=4$ (or vice versa), so $H_2$ is abelian, and hence $|Z(H_1 times H_2)| =8$. But you can check that $Z(H)=Z(G) = langle a_1^2,a_2^2 rangle$ has order $4$.
In fact $H$ is indecomposable.
answered Nov 22 at 14:10
Derek Holt
52.3k53570
Let $G=D_8 = langle a,b mid a^4=b^2=(ab)^2=1 rangle$, $G^2 = langle a_1,b_1rangle times langle a_2,b_2 rangle$ and $H = langle a_1a_2,b_1,b_2,a_1^2,a_2^2 rangle le G$ with $|H|=32$.
If $H cong H_1 times H_2$, with $H_1$ and $H_2$ isomorphic to subgroups of $D_8$, then we must have $|H_1|=8$, $|H_2|=4$ (or vice versa), so $H_2$ is abelian, and hence $|Z(H_1 times H_2)| =8$. But you can check that $Z(H)=Z(G) = langle a_1^2,a_2^2 rangle$ has order $4$.
In fact $H$ is indecomposable.
answered Nov 22 at 14:10
Derek Holt
52.3k53570
answered Nov 22 at 14:10
Derek Holt
52.3k53570
answered Nov 22 at 14:10
Derek Holt
52.3k53570
answered Nov 22 at 14:10
Derek Holt
52.3k53570
52.3k53570
add a comment |
add a comment |
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Have you run any computer checks for small values of $|G|$ and $n$?
– the_fox
Nov 22 at 12:25
I tried a few examples on GAP without finding any counterexample so far.
– J.-E. Pin
Nov 22 at 12:34
I doubt this is true, but I could be wrong. Include the code you have used to check this, if you want, and we can split the workload. (I was just about to start thinking how to write a piece of GAP code that would work, so it would save me some time.)
– the_fox
Nov 22 at 12:41
I have replaced "Is any" in your question by "Are all". The word "any" is very often ambiguous.
– Derek Holt
Nov 22 at 12:49
1
I think $D_8 times D_8$ has a subdirect product of index 2 that does not decompose as a nontrivial direct product. (Here $D_8$ means dihedral group of order $8$.)
– Derek Holt
Nov 22 at 12:55