To finde the center of $D_4$

Question

is there a nice/smart way to find the center of $D_4$? rather then going through every element?

Answer 1

I think you can use the following fact:

Let $G$ be a finite group with center $Z(G)$, if the quotient $G/Z(G)$ is cyclic, then $G$ is abelian.

We obtain that the only possible cardinalities for $Z(D_4)$ are 2 or 1.

Now, you have only to show that there's a non trivial element such that commutes with all the other ones in $D_4$, but this is very easy (Hint: if $\rho$ is such that $\rho^4=e$, look at $\rho^2$).

Answer 2

By its represantation, $$D_4=<a,b> | bab^{-1}=a^{-1} , a^4=e=b^2$$

Thus, $ba^2b^{-1}=a^{-2}=a^2$ so $a^2\in Z(D_4)$

From that we see that $Z(G)$ has at least $2$ elements but can not be $4$ as $G/Z(G)$ become cylic and can not be $8$ as it is nonabelian.

$$Z(D_4)=\{e,a^2\}$$