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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 $C_GC_G(H)$

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  • $\begingroup$ 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 $C_G ( C_G (H))$. $\endgroup$ – Ashish K Oct 25 '18 at 18:39
  • $\begingroup$ ah okay I think I get you . I'll try writing that proof out now. $\endgroup$ – excalibirr Oct 25 '18 at 18:41
  • $\begingroup$ @AshishK I wrote a new answer , is it now correct ? $\endgroup$ – excalibirr Oct 25 '18 at 19:09
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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 $H\leq C_G(C_G(H))$

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  • $\begingroup$ 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 $h \in C_G (C_G (H))$. $\endgroup$ – Ashish K Oct 25 '18 at 19:28
  • $\begingroup$ And I am not sure about your @exodius last sentence which seems to be nonsense $\endgroup$ – Nicky Hekster Oct 25 '18 at 20:44
  • $\begingroup$ I made a mistake , I fixed it now though $\endgroup$ – excalibirr Oct 25 '18 at 20:45

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