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

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

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.

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

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

3. Originally Posted by Swlabr
Where is it you are stuck on this question?

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

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

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...

4. 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 $\displaystyle H$ is a sugroup of $\displaystyle G$ you have to prove two things,

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

5. Originally Posted by Swlabr
It'll be in your notes. To prove that $\displaystyle H$ is a sugroup of $\displaystyle G$ you have to prove two things,

That for all $\displaystyle g, h \in H$ then $\displaystyle g*h \in H$, and for all $\displaystyle h \in H$, $\displaystyle h^{-1} \in H$.
And you'll also need to show there's at least something in the group, usually $\displaystyle e\in H$ suffices. The reason is that the empty set satisfies those two conditions but it's not a group