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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.

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  • $\begingroup$ 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
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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!!!

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  • $\begingroup$ 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

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