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Let $H\leq G$. Show that $H$ is normal iff $xHx^{-1}=H\space \forall x\in G$.

My textbook defines normal subgroup of $G$ as kernel of some homomorphism which has $G$ as domain.

I showed that if $H$ is normal, than $xHx^{-1}=H$ holds. Suppose $f$ is some homomorphism from $G$ with $H$ in the kernel. Consider $h\in H$. Then $f(xhx^{-1})=f(x)f(h)f(x^{-1})=f(x)f(x)^{-1}=e'$. So $xhx^{-1}$ lies in the kernel and $xHx^{-1}=H$.

Can you give me idea how to show that if $xHx^{-1}=H$ then $H$ is a kernel of some homomorphism from $G$?

Mihail
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  • I think an easy way would be to create such a homomorphism. – Leppala Jan 23 '15 at 10:03
  • You don't know about quotient groups yet, do you? – user208259 Jan 23 '15 at 10:04
  • @user208259 No. I know only a little bit of homomorphisms and cosets with normal subgroups. – Mihail Jan 23 '15 at 10:05
  • Did you now that when $N$ is normal in $G$ then the set of $N$-cosets (either left or right) has the structure of a group? This is the "quotient group". – Ofir Schnabel Jan 23 '15 at 10:08
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    Well, the ordinary way to do this is to define a group structure on the set of cosets. Define the product of two subsets $S$ and $T$ of $G$ to be $ST = {st \mid s \in S, t \in T }$. Now use this rule to define the "product" of two cosets. On the set of cosets $G/H$, check that this defines a group structure. Then consider the morphism $g \mapsto gH$. – user208259 Jan 23 '15 at 10:09
  • Knowing cosets and that $G/N$ is a group should be enough. – Leppala Jan 23 '15 at 10:09
  • My textbook calls it factor group $G/H$. Do you mean this? – Mihail Jan 23 '15 at 10:09
  • Think to the map $\varphi: G \to G/H, g\mapsto gH$, with $H\lhd G$. Is she well defined? What is the kernel of this map? – Bman72 Jan 23 '15 at 10:10
  • Yes, a quotient group is sometimes called a factor group. – Leppala Jan 23 '15 at 10:11
  • See also http://math.stackexchange.com/a/1014784/589. – lhf Jan 23 '15 at 10:44

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Thanks for comments. Based on them I'll try to answer on my own question. $xHx^{-1}=H \implies xH=Hx$. So we can consider a quotient group $G/H$. Now consider a homomorphism $f:G\to G/H$, where $f(x)=xH$. The kernel is $\{x\in G: f(x)=eH=H\}$. I have to show that this set is equivalent to $H$. Since $H$ is a group then $xH$ for $x\in H$ is a subset of $H$. If $x\in G$ but $x\notin H$ then $xh\notin H$ just because $xe\notin H$. So $H=\{x\in G: f(x)=eH=H\}$ which means that $H$ is a kernel or normal subgroup of $G$.

Is there something wrong?

Mihail
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    As far as i can see no. Maybe you should show that the map is well defined, i.e. it does not depend on the choice of the representant pf the equivalence class: if $a'=ah_1, b'=bh_2,$ then $a'b'=ah_1bh_2=abh_1'h_2=abH$, because $H\lhd G \implies Hb=bH$. – Bman72 Jan 23 '15 at 10:35