Prove that for any a E G, Ca is a subgroup of G.

**rainyice** · Apr 2nd 2010, 09:52 PM

For a fixed element a of a group G, the set Ca = {x E G | ax = xa } is the centralizer of a in G.

Prove that for any a E G, Ca is a subgroup of G.

**Swlabr** · Apr 3rd 2010, 12:08 AM

Originally Posted by rainyice

For a fixed element a of a group G, the set Ca = {x E G | ax = xa } is the centralizer of a in G.

Prove that for any a E G, Ca is a subgroup of G.

Where is it you are stuck on this question?

Let $x, y \in C_a$ . Then you need to check that $xy \in Ca$ and $x^{-1} \in Ca$ . This is precisely what you would always do to check something is a subgroup.

Checking that $xy$ is in $C_a$ is almost elementary (substitute $`xy'$ in the above condition for $`x$ ), but the fact that $x^{-1} \in C_a$ is not immediately obvious. However, just take the condition, and pre- and post-multiply by $x^{-1}$ to get that the condition also holds for $x^{-1}$

**rainyice** · Apr 3rd 2010, 02:09 PM

Originally Posted by Swlabr

Where is it you are stuck on this question?

Let $x, y \in C_a$ . Then you need to check that $xy \in Ca$ and $x^{-1} \in Ca$ . This is precisely what you would always do to check something is a subgroup.

Checking that $xy$ is in $C_a$ is almost elementary (substitute $`xy'$ in the above condition for $`x$ ), but the fact that $x^{-1} \in C_a$ is not immediately obvious. However, just take the condition, and pre- and post-multiply by $x^{-1}$ to get that the condition also holds for $x^{-1}$

Currently, this is my first time start to know Group Theory. I don't really know how to prove if it is a subgroup or not because I do not know where and how I should start...

**Swlabr** · Apr 4th 2010, 08:25 AM

Originally Posted by rainyice

Currently, this is my first time start to know Group Theory. I don't really know how to prove if it is a subgroup or not because I do not know where and how I should start...

It'll be in your notes. To prove that $H$ is a sugroup of $G$ you have to prove two things,

That for all $g, h \in H$ then $g*h \in H$ , and for all $h \in H$ , $h^{-1} \in H$ .

**FancyMouse** · Apr 4th 2010, 09:31 AM

Originally Posted by Swlabr

It'll be in your notes. To prove that $H$ is a sugroup of $G$ you have to prove two things,

That for all $g, h \in H$ then $g*h \in H$ , and for all $h \in H$ , $h^{-1} \in H$ .

And you'll also need to show there's at least something in the group, usually $e\in H$ suffices. The reason is that the empty set satisfies those two conditions but it's not a group

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