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:


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


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


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


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.

This entry was posted in Algebra, Centralizer, Normal. Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s