problem: For a subgroup $N$ of a group $G$, prove if $N$ is normal, then each left coset of $N$ is also a right coset, that is for all $a \in G$, there exists a $b \in G$, such that $aN = Nb$.

Attempt: Suppose $N$ is a normal subgroup. Then we know $N = aNa^{-1}$, for all $a \in G$. Suppose $ana^{-1}$ is an element in $N$, then there exists a $b \in N$ such that $b = ana^{-1}$. So multiply both sides of $b = ana^{-1}$ by $a$, to get $ba = an$. Then if $c$ is an element in $N$, then $ba = an$ implies $(ba)c = (an)c = a(nc)$. So $a(nc)$ is in $N$. Since both $n$ and $c$ are also in $N$.

Similarly $N = bnb^{-1}$. Suppose $bnb^{-1}$ is an element in $N$, then there exists an $a$ such that $a = bnb^{-1}$. Thus $ab = bn$. So if $c$ is an element in $N$, then $ab = bn$ implies $c(bn) = c(ab) = (ca)b$ which is an element of $Nb$. So $aN \subset Nb$ and $Nb \subset aN$. Thus $aN = Nb$.

Can someone please check if this does make sense? And please any help/feedback/hints, would be really appreciated. Thank you.

  • $\begingroup$ First line: Careful. We know only that, for some $a \in G$, $aNa^{-1} \subset N$. $\endgroup$ – Kaj Hansen Apr 1 '14 at 5:22
  • $\begingroup$ @Kaj_H: if $N$ if normal in $G$, then $aNa^{-1} = N$ for all $a \in G$, if I am not mistaken. That is the definition of normal subgroup. $\endgroup$ – Robert Lewis Apr 1 '14 at 5:32
  • $\begingroup$ @RobertLewis: You're right. Oops. $\endgroup$ – Kaj Hansen Apr 1 '14 at 5:35
  • $\begingroup$ @Kaj_H: these definitions get so confusing, especially when one is learning them, and there are so many of them, it is easy to make mistakes. After my algebra final I got them so mixed up it took me years to unravel this stuff! ;-)! Cheers! $\endgroup$ – Robert Lewis Apr 1 '14 at 5:51

I'm having a hard time following the argument presented in the text of the question, and I don't think it is completely correct. For example, since $nc \in N$, if $a \in G - N$ (that is, $a \in G$, $a \notin N$), then $a(nc) \notin N$. If it were, say $a(nc) = m \in N$, then $a = m(nc)^{-1} \in N$, a contradiction. In fact, $a(nc) \in aN$, a left coset of $N$, and $N \cap aN = \varnothing$ since left cosets are either disjoint or identical, as are right cosets.

The easiest way I know to show that left cosets are right cosets (and vice versa!) for normal $N$ is to use $aNa^{-1} = N$; then $aN = Na$ so the left and right cosets represented by $a$ (that is, containing $a$) are identical. It's as simple as that.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

| cite | improve this answer | |
  • $\begingroup$ I understand it that way, but that b in the problem is getting me confuse. $\endgroup$ – user65903 Apr 1 '14 at 5:55
  • $\begingroup$ @user65903: if the problem is to show that $b$ exists, just take $b = a$ for any $a \in G$! $\endgroup$ – Robert Lewis Apr 1 '14 at 5:57
  • $\begingroup$ Thank you, then I will do that. $\endgroup$ – user65903 Apr 1 '14 at 6:00
  • $\begingroup$ @user65903: (continuation of previous comment) This exhibits a $b$ with the desired property $aG = Gb (= Ga!)$; you don't need to manipulate $b$ other than this! $\endgroup$ – Robert Lewis Apr 1 '14 at 6:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.