# Computing a centralizer in D_8

## Homework Statement

If $A = \{1,r^2,s,sr^2 \}$, compute $C_{D_8}(A)$

## The Attempt at a Solution

So I am not exactly sure what is the best way to do this, but here is what I tried. From previous problems I know that $Z(D_8) = \{1, r^2 \}$. So $Z(G) \le C_{D_8}(A) \le D_8$, so the order of $C_{D_8}(A)$ can only be 2, 4, or 8. It's not 8 since we can compute directly that $r \not \in C_{D_8}(A)$. It's not 2 since we can compute directly that $s \in C_{D_8}(A)$. So the order is 4. And since by computation we see that $sr^2 \in C_{D_8}(A)$, we have that $C_{D_8}(A) = A$.

## Answers and Replies

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fresh_42
Mentor
How to read the index? I only know the notation $|D_n|=2n$ but I assume you mean $|D_8|=8$.

How to read the index? I only know the notation $|D_n|=2n$ but I assume you mean $|D_8|=8$.
I am using the notation that $|D_{2n}| = 2n$, which is what's used in Dummit and Foote.

fresh_42
Mentor

## Homework Statement

If $A = \{1,r^2,s,sr^2 \}$, compute $C_{D_8}(A)$

## The Attempt at a Solution

So I am not exactly sure what is the best way to do this, but here is what I tried. From previous problems I know that $Z(D_8) = \{1, r^2 \}$. So $Z(G) \le C_{D_8}(A) \le D_8$, so the order of $C_{D_8}(A)$ can only be 2, 4, or 8. It's not 8 since we can compute directly that $r \not \in C_{D_8}(A)$. It's not 2 since we can compute directly that $s \in C_{D_8}(A)$. So the order is 4. And since by computation we see that $sr^2 \in C_{D_8}(A)$, we have that $C_{D_8}(A) = A$.
Maybe correct, maybe not. You haven't done the multiplications, and I'm too lazy to do it again.

What I have observed is, that $A$ is the Klein four-group $V_4 \cong \mathbb{Z}_2^2$. Since $D_8$ (I really don't like this notation) isn't Abelian and $V_4$ normal in $D_8$, we have $D_8 \cong V_4 \rtimes \mathbb{Z}_2 = A \rtimes \mathbb{Z}_2$. Since $V_4$ is Abelian, we have $D_8 \supsetneq C_{D_8}(A) \supseteq A$ and therefore by Lagrange $C_{D_8}(A)=A$.