Abstract algebra












0














Question




Prove that $a=a^2$ in a group if and only if $a=e$




Attempt
Suppose that $a^2=a$ such that $ain G$ , that is $(G, *)$
Then $a*e=e*a=a$
Implies that $ae=ea=a$
Now suppose that $a≠e$
Since $a^{-1}a=aa^{-1}=e$
Then $a(aa^{-1})=e$
$aa=e$
$a^2=e$
Now suppose that $a=e$
Then $a=a^2$



Please help me with your solution so that I would know that what I have done is correct.






group-theory proof-verification







share|cite|improve this question
























  • The proof looks very circular and doesn't actually make the conclusion that you want. You have unnecessary steps (you say that $ae = a$ implies $ae=a$...?), you don't use the assumption that $a ne e$, and you have at least one incorrect conclusion. After the step $a(aa^{-1}) = e$ (which is wrong), you get $aa = e$, which doesn't follow. So I don't really see that you've ever used an assumption that $a = a^2$. For a successful proof, perhaps assume $a = a^2$ and multiply both sides by $a^{-1}$.
    – T. Bongers
    Nov 26 '18 at 19:53










  • You can add mathjax to improve your post. See the edits to learn how this can be done.
    – Mason
    Nov 26 '18 at 20:08










  • Thanks bh can you at least help me with a brief intro
    – Fosu Emmanuel
    Nov 29 '18 at 18:49
















0














Question




Prove that $a=a^2$ in a group if and only if $a=e$




Attempt
Suppose that $a^2=a$ such that $ain G$ , that is $(G, *)$
Then $a*e=e*a=a$
Implies that $ae=ea=a$
Now suppose that $a≠e$
Since $a^{-1}a=aa^{-1}=e$
Then $a(aa^{-1})=e$
$aa=e$
$a^2=e$
Now suppose that $a=e$
Then $a=a^2$



Please help me with your solution so that I would know that what I have done is correct.






group-theory proof-verification







share|cite|improve this question
























  • The proof looks very circular and doesn't actually make the conclusion that you want. You have unnecessary steps (you say that $ae = a$ implies $ae=a$...?), you don't use the assumption that $a ne e$, and you have at least one incorrect conclusion. After the step $a(aa^{-1}) = e$ (which is wrong), you get $aa = e$, which doesn't follow. So I don't really see that you've ever used an assumption that $a = a^2$. For a successful proof, perhaps assume $a = a^2$ and multiply both sides by $a^{-1}$.
    – T. Bongers
    Nov 26 '18 at 19:53










  • You can add mathjax to improve your post. See the edits to learn how this can be done.
    – Mason
    Nov 26 '18 at 20:08










  • Thanks bh can you at least help me with a brief intro
    – Fosu Emmanuel
    Nov 29 '18 at 18:49














0












0








0







Question




Prove that $a=a^2$ in a group if and only if $a=e$




Attempt
Suppose that $a^2=a$ such that $ain G$ , that is $(G, *)$
Then $a*e=e*a=a$
Implies that $ae=ea=a$
Now suppose that $a≠e$
Since $a^{-1}a=aa^{-1}=e$
Then $a(aa^{-1})=e$
$aa=e$
$a^2=e$
Now suppose that $a=e$
Then $a=a^2$



Please help me with your solution so that I would know that what I have done is correct.






group-theory proof-verification







share|cite|improve this question















Question




Prove that $a=a^2$ in a group if and only if $a=e$




Attempt
Suppose that $a^2=a$ such that $ain G$ , that is $(G, *)$
Then $a*e=e*a=a$
Implies that $ae=ea=a$
Now suppose that $a≠e$
Since $a^{-1}a=aa^{-1}=e$
Then $a(aa^{-1})=e$
$aa=e$
$a^2=e$
Now suppose that $a=e$
Then $a=a^2$



Please help me with your solution so that I would know that what I have done is correct.






group-theory proof-verification




group-theory proof-verification






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 26 '18 at 20:26









Mason

1,9591530




1,9591530










asked Nov 26 '18 at 19:49









Fosu Emmanuel

31




31












  • The proof looks very circular and doesn't actually make the conclusion that you want. You have unnecessary steps (you say that $ae = a$ implies $ae=a$...?), you don't use the assumption that $a ne e$, and you have at least one incorrect conclusion. After the step $a(aa^{-1}) = e$ (which is wrong), you get $aa = e$, which doesn't follow. So I don't really see that you've ever used an assumption that $a = a^2$. For a successful proof, perhaps assume $a = a^2$ and multiply both sides by $a^{-1}$.
    – T. Bongers
    Nov 26 '18 at 19:53










  • You can add mathjax to improve your post. See the edits to learn how this can be done.
    – Mason
    Nov 26 '18 at 20:08










  • Thanks bh can you at least help me with a brief intro
    – Fosu Emmanuel
    Nov 29 '18 at 18:49


















  • The proof looks very circular and doesn't actually make the conclusion that you want. You have unnecessary steps (you say that $ae = a$ implies $ae=a$...?), you don't use the assumption that $a ne e$, and you have at least one incorrect conclusion. After the step $a(aa^{-1}) = e$ (which is wrong), you get $aa = e$, which doesn't follow. So I don't really see that you've ever used an assumption that $a = a^2$. For a successful proof, perhaps assume $a = a^2$ and multiply both sides by $a^{-1}$.
    – T. Bongers
    Nov 26 '18 at 19:53










  • You can add mathjax to improve your post. See the edits to learn how this can be done.
    – Mason
    Nov 26 '18 at 20:08










  • Thanks bh can you at least help me with a brief intro
    – Fosu Emmanuel
    Nov 29 '18 at 18:49
















The proof looks very circular and doesn't actually make the conclusion that you want. You have unnecessary steps (you say that $ae = a$ implies $ae=a$...?), you don't use the assumption that $a ne e$, and you have at least one incorrect conclusion. After the step $a(aa^{-1}) = e$ (which is wrong), you get $aa = e$, which doesn't follow. So I don't really see that you've ever used an assumption that $a = a^2$. For a successful proof, perhaps assume $a = a^2$ and multiply both sides by $a^{-1}$.
– T. Bongers
Nov 26 '18 at 19:53




The proof looks very circular and doesn't actually make the conclusion that you want. You have unnecessary steps (you say that $ae = a$ implies $ae=a$...?), you don't use the assumption that $a ne e$, and you have at least one incorrect conclusion. After the step $a(aa^{-1}) = e$ (which is wrong), you get $aa = e$, which doesn't follow. So I don't really see that you've ever used an assumption that $a = a^2$. For a successful proof, perhaps assume $a = a^2$ and multiply both sides by $a^{-1}$.
– T. Bongers
Nov 26 '18 at 19:53












You can add mathjax to improve your post. See the edits to learn how this can be done.
– Mason
Nov 26 '18 at 20:08




You can add mathjax to improve your post. See the edits to learn how this can be done.
– Mason
Nov 26 '18 at 20:08












Thanks bh can you at least help me with a brief intro
– Fosu Emmanuel
Nov 29 '18 at 18:49




Thanks bh can you at least help me with a brief intro
– Fosu Emmanuel
Nov 29 '18 at 18:49










1 Answer
1






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oldest

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2














In every group $G$ we have the cancellation law. So we can "cancel" an $a$, by multiplying with $a^{-1}$ on both sides of the equation $a=a^2=aa$, to obtain that $a^{-1}a=a^{-1}aa$, which says that $e=a$.






share|cite|improve this answer





















  • So can I start by answering it by: supposing that aa^-1£G. Since a^-1 is an inverse element in a we have that aa^-1=a^-1a=e
    – Fosu Emmanuel
    Nov 29 '18 at 19:02













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1 Answer
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1 Answer
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active

oldest

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active

oldest

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active

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votes









2














In every group $G$ we have the cancellation law. So we can "cancel" an $a$, by multiplying with $a^{-1}$ on both sides of the equation $a=a^2=aa$, to obtain that $a^{-1}a=a^{-1}aa$, which says that $e=a$.






share|cite|improve this answer





















  • So can I start by answering it by: supposing that aa^-1£G. Since a^-1 is an inverse element in a we have that aa^-1=a^-1a=e
    – Fosu Emmanuel
    Nov 29 '18 at 19:02


















2














In every group $G$ we have the cancellation law. So we can "cancel" an $a$, by multiplying with $a^{-1}$ on both sides of the equation $a=a^2=aa$, to obtain that $a^{-1}a=a^{-1}aa$, which says that $e=a$.






share|cite|improve this answer





















  • So can I start by answering it by: supposing that aa^-1£G. Since a^-1 is an inverse element in a we have that aa^-1=a^-1a=e
    – Fosu Emmanuel
    Nov 29 '18 at 19:02
















2












2








2






In every group $G$ we have the cancellation law. So we can "cancel" an $a$, by multiplying with $a^{-1}$ on both sides of the equation $a=a^2=aa$, to obtain that $a^{-1}a=a^{-1}aa$, which says that $e=a$.






share|cite|improve this answer












In every group $G$ we have the cancellation law. So we can "cancel" an $a$, by multiplying with $a^{-1}$ on both sides of the equation $a=a^2=aa$, to obtain that $a^{-1}a=a^{-1}aa$, which says that $e=a$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 26 '18 at 19:56









Dietrich Burde

77.6k64386




77.6k64386












  • So can I start by answering it by: supposing that aa^-1£G. Since a^-1 is an inverse element in a we have that aa^-1=a^-1a=e
    – Fosu Emmanuel
    Nov 29 '18 at 19:02




















  • So can I start by answering it by: supposing that aa^-1£G. Since a^-1 is an inverse element in a we have that aa^-1=a^-1a=e
    – Fosu Emmanuel
    Nov 29 '18 at 19:02


















So can I start by answering it by: supposing that aa^-1£G. Since a^-1 is an inverse element in a we have that aa^-1=a^-1a=e
– Fosu Emmanuel
Nov 29 '18 at 19:02






So can I start by answering it by: supposing that aa^-1£G. Since a^-1 is an inverse element in a we have that aa^-1=a^-1a=e
– Fosu Emmanuel
Nov 29 '18 at 19:02




















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