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$\begingroup$

I am in the process of proving that $C_{G}(H)$ is a normal subgroup of $N_{G}(H)$, where $C_{G}(H)$ is the centralizer - i.e., the set of elements $g \in G$ that commute with all $h \in H \leqslant G$ - and $N_{G}(H)$ is the normalizer - i.e., the set of all $g \in G$ such that $g^{-1}Hg = H$.

It is clear that $C_{G}(H) \subset N_{G}(H)$, but before I can show that $C_{G}(H)$ is a normal subgroup of $N_{G}(H)$, I should probably show that it is just a subgroup of it first.

In this regard, I have not been able to find much help online. Every proof I've seen has kind of glazed over it as a trivial detail, but I do not have that luxury. As part of my proof that $C_{G}(H)$ is a normal subgroup of $N_{G}(H)$, I must explicitly show this first.

To show that $C_{G}(H)$ is a subgroup of $H$ is easy - it contains the identity, is closed under multiplication and under taking inverses. But, and I might be overthinking this - is it different to show that it is a subgroup of $N_{G}(H)$? In other words, what should the operation of multiplication look like in this case? Should I try to show that $(zw)^{-1}H(zw)=H$, where $z,w \in C_{G}(H)$? And what would taking inverses look like?

Again, I am only asking about showing that $C_{G}(H) \leqslant N_{G}(H)$. NOT about the whole proof of showing that $C_{G}(H) \trianglelefteq N_{G}(H)$.

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    You have to show every element of $C_G(H)$ is in $N_G(H)$. Can you do that? (Unpack the definitions: $z$ being an element of $C_G(H)$ means ___, and $z$ being an element of $N_G(H)$ means ____.) – 2017-01-01
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    If you have proved the $N(H)$ is a group and $C(H)\subset N(H)$ it is same as you prove for $C(H)\subset G$. Re-read one step/two step subgroup test or separately check closure identity inverese and associativity – 2017-01-01
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    @Bhaskar What? ${}$ – 2017-01-01
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    @arctictern They mention in the 2nd paragraph that they've shown that inclusion. I think they're just unsure about whether it's sufficient to claim that the centralizer is a subgroup of the normalizer. – 2017-01-01
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    Try to prove "Let G be a group and H is a subgroup of G. Then H is also a subgroup of any group K such that $H\subset K – 2017-01-01
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    @BhaskarVashishth you're essentially telling me to prove the very thing I'm asking how to prove. I don't know how the mechanics should work in this case. – 2017-01-01
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    http://math.stackexchange.com/questions/1546650/show-the-centralizer-of-h-in-g-is-a-subgroup-of-the-normalizer-of-h-in-g – 2017-01-01
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    @BhaskarVashishth thanks. I also just wrote out what you suggested above that I said was what I was asking to prove and proved it - it was really quite silly and easily done. I am the queen of overthinking. – 2017-01-01
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    If it's a subset of a subgroup and it is a subgroup of the entire group, then it is a subgroup of the subgroup. The subgroup relation is pretty simple. Subtlety arises when you start asking about normality. – 2017-01-02
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    Overthinking is in math much better than underthinking. Stay sceptical of every step in every proof! – 2017-01-02

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