## Gallian- Ch.9 #49

Problem Statement: If $H\lhd G$, prove that $C(H)\lhd G$.

Proof: Recall that $C(H)$ is the set of all elements of $G$ which commute with all elements of $H$. Let $g\in G$, we wish to show that $gC(H)c^{-1}\subset C(H)$. So consider an element $s\in gC(H)g^{-1}$. Then there is some $x\in C(H)$ such that $s=gxg^{-1}$. Now, in order to show that $s\in C(H)$ we must show that $s$ commutes with any element $h\in H$.

Consider $gxg^{-1}h$. Since $H\lhd G$ it follows that there exists some $h'\in H$ such that $g^{-1}h=h'g^{-1}$.  Note that this also means that $gh'=hg$. Using these facts we may rewrite the expression in the following way:

$gxg^{-1}h=gxh'g^{-1}$

But $x$ commutes with everything from $H$ and so we may again rewrite the expression:

$=gh'xg^{-1}$

But as stated above, $gh'=hg$ and so we may rewrite one more time:

$=hgxg^{-1}$

And so we may conclude that $gC(H)g^{-1}\subset C(H)$ for every $g\in G$ and so $C(H)\lhd G$.

$\Box$

Reflection: This is one of those proofs that is not entirely difficult, but is very easy to make much more difficult than it needs to be. Thankfully I have Jeremy to straighten me out when I over think these simple problems. The key thing for me while doing this proof was to think in words, and less in symbols. Once we started talking about what it means to be in the centralizer of $H$ then it started to click.