# Help understanding what's needed in this proof to show that H is a subgroup.

I want to show that given that $$H \leq G$$ that we also have $$H \leq C_G(C_G(H)).$$

My attempt :

i think the best way to do this is break it down into stages so :

Stage 1) showing that H is a subset of $$C_G(C_G(H))$$

$$C_G(C_G(H)):=\{g\in G |cg=gc, \forall c \in C_G(H)\}$$.

which is the set containing all the elements of G which commute with all the elements of $$C_G(H)$$. But $$C_G(H)$$ is the set set which contains all the elements of G which commute with all elements of H. Which means that $$C_G(C_G(H)$$ is the set containing all the elements of G which commute with the elements of G that commute with all the elements of h.

well h satisfies the property of of commuting with h which commutes with h. therefore H must be contained in $$C_G(C_G(H))$$.

Stage 2) showing that for all $$a,b \in H,ab^{-1}\in H$$, Although I'm not sure how to proceed from here. Any suggestions ?

I think maybe we could say that as all elements in h are of the form

(hch^{-1}gh^{-1}ch)(h^{-1}c^{-1}h g^{-1}hc^{-1}h^{-1}

then by cancelation we get that this is equal to the identity which is in H as it is a subgroup meaning that it is a subgroup of $$C_GC_G(H)$$

• It has been given that $H$ is a subgroup of $G$. You proved that $H$ is a subset of $C_G ( C_G (H))$. If you can prove that $C_G ( C_G (H))$ is a subgroup of $G$, then you will have proven that $H$ is a subgroup of $C_G ( C_G (H))$. – Ashish K Oct 25 '18 at 18:39
• ah okay I think I get you . I'll try writing that proof out now. – excalibirr Oct 25 '18 at 18:41
• @AshishK I wrote a new answer , is it now correct ? – excalibirr Oct 25 '18 at 19:09

First we prove that $$C_G(C_G(H)) \leq G$$. Well We know that $$C_G(H)\leq G \Rightarrow C_G(C_G(H))\leq G$$.

We already know that $$H\leq G$$

So now all we need to show is that

$$H\subset C_G(C_G(H))$$

Well let $$g\in C_G(C_G(H)$$

$$\Rightarrow g=(hch^{-1})g(hch^{-1})^{-1}=(hch^{-1})g(h^{-1}c^{-1}h)$$.

Now suppose that g=h , now this equation is obviously true

$$hch^{-1}hh^{-1}c^{-1}h=hch^{-1}c^{-1}h=hc(hc)^{-1}h=h.$$ and so the condition to be a member of $$C_G(C_G(H))$$ is met by h therefore $$H\subset C_G(C_G(H))$$.

Now we use the fact that $$|H|<|C_G(C_G(H))|$$ and $$H,C_G(C_G(H))\leq G$$ to assume that $$H\leq C_G(C_G(H))$$

• I'm not sure how you concluded that $H \subseteq C_G (C_G (H))$. To prove this, you need to take an arbitrary $h \in H$ and show that $h \in C_G (C_G (H))$. – Ashish K Oct 25 '18 at 19:28
• And I am not sure about your @exodius last sentence which seems to be nonsense – Nicky Hekster Oct 25 '18 at 20:44
• I made a mistake , I fixed it now though – excalibirr Oct 25 '18 at 20:45