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For a group $G$ and a normal subgroup $N$ of $G$, the quotient group of $N$ in $G$, written $G/N$ and read "$G$ modulo $N$", is the set of cosets of $N$ in $G$.

Question : Why in the defintion of Quotient Group we need subgroup $N$ to be normal?

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marked as duplicate by rschwieb, Lord_Farin, Martin R, Martin Sleziak, Lord Shark the Unknown Jan 23 at 3:54

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    $\begingroup$ This is used to show that the product $(gN)(hN)=(gh)N$ is well defined. $\endgroup$ – John B Feb 17 '18 at 13:10
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    $\begingroup$ Try to define a similar quotient structure without assuming normality and see what the problems are. $\endgroup$ – the_fox Feb 17 '18 at 13:10
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Let $G$ be a group and let $e$ denote its identity.

Forming a quotient $G/\sim$ of group $G$ is actually forming a partition on $G$ such that $[a][b]=[ab]$ is a well defined multiplication. Here $[a]$ and $[b]$ are elements of the partition that are represented by $a\in[a]$ and $b\in[b]$ respectively.

The multiplication is well defined if $[a'b']=[ab]$ whenever $a'\in[a]$ and $b'\in[b]$. Further it is not difficult to prove that $[e]$ will be a subgroup and will work as identity of the quotient.

Now if $a\in[e]$ and $g\in G$ then we find $$[gag^{-1}]=[g][a][g^{-1}]=[g][e][g^{-1}]=[geg^{-1}]=[e]$$This proves the necessity of: $$a\in[e]\implies gag^{-1}\in[e]$$ So $[e]$ is bound to be a normal subgroup.

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We define the multiplication of two cosets $gN$ and $hN$ to be $ghN$. So, if $N$ is not a normal subgroup of $G$, then this multiplication will not be well defined, i.e. there may be two different and non-equal values for $(gN)(hN)$.

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  • $\begingroup$ @shi, take the smallest possible example: $G=S_3$ and $H=\{e, (12)\}$ with $g=h=(123)$. $\endgroup$ – ancientmathematician Feb 17 '18 at 13:41
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That is to have a well-defined group law on the quotient: for $a,b\in G$, we need to check that , if $a'\equiv a\mod H$, $b'\equiv b\mod H$*, we want that $a'b'\equiv ab\mod H$.

Explicitly, this means that, if $a'=ah$, $\;b'=bk\;$ for some $h,k\in H$, we must have $$a'b'(=ah\mkern1.5mubk)=ab \mkern1.5mu\ell\quad\text{for some }\enspace \ell\in H.$$ In particular, we must have $$a'a^{-1}=aha^{-1}\in H\enspace\text{for any }h\in H,~a\in G, $$ which exactly means $H$ is normal in $G$.

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Just as an example: take the (not normal) subgroup $\{i,(23)\}=H\subset S_3$, where $i$ is the identity. Indeed, $$H_{(132)}=\{(132),(12)\}=H_{(12)}$$ and $$H_{(123)}=\{ (123),(13)\}=H_{(13)}.$$ However, $H_{(132)\circ (123)}=H_i$, which is different from $H_{(12)\circ (13)}=H_{(132)}$, so that our operation isn't well-defined.

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