# Computing a centralizer in D_8

• Mr Davis 97

## 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##.

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.

## 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##.

• Mr Davis 97