If G ~ H, Z(G) ~ Z(H)

[Mr Davis 97]

Homework Statement

Prove that if the groups $G \cong H$ are isomorphic then $Z(G) \cong Z(H)$

Homework Equations

The Attempt at a Solution

Let $\phi: G \to H$ be an isomorphism. Define $f: Z(G) \to Z(H)$ s.t $f(z) = \phi (z)$.
First, we will show that this map is well-defined, in the sense that elements in $Z(G)$ are always mapped to elements in $Z(H)$ by $f$:

Let $z \in Z(G)$. Then $zg=gz$ for all $g \in G$. This implies that $f(zg)=f(gz) \implies \phi (z) \phi (g) = \phi (g) \phi (z)$ for all $g$. But $g$ is arbitrary and $\phi$ is bijective, so $\phi (g)$ is an arbitrary element of $H$. Hence we have $\phi(z)h=h \phi(z)$ for all $h \in H$, which means $f(z)h=h f(z)$ for all $h \in H$. So $f(z) \in Z(H)$.

Next, we show that $f$ is an isomorphism: It is clear that $f$ is a homomorphism, since it is defined in terms of $\phi$. It remains to show that $f$ is a bijection: $f$ is invertible, and the inverse is $f^{-1} = \phi^{-1}$.

Hence $Z(G) \cong Z(H)$

 

[fresh_42]

Correct. The bijection part is a bit short, e.g. you could have added that $f^{-1}$ which isn't defined yet, is into $Z(G)$ for the same reason as $f$ was, that is that $\phi^{-1}$ can be restricted to the centers, too.

And I won't say that this is a matter of ... I basically have written this in order to say at least something besides "correct".

 

[Mr Davis 97]

Correct. The bijection part is a bit short, e.g. you could have added that $f^{-1}$ which isn't defined yet, is into $Z(G)$ for the same reason as $f$ was, that is that $\phi^{-1}$ can be restricted to the centers, too.

And I won't say that this is a matter of ... I basically have written this in order to say at least something besides "correct".

I have one question that's slightly related to this one, and one which I don't want to create a new thread for. In trying to show that $G \cong H \implies \operatorname{Aut} (G) \cong \operatorname{Aut} (H)$, one supposes that $\phi : G \to H$ and considers the map $\theta:\text{Aut}(G)\rightarrow\text{Aut}(H)$ defined as $\theta(\psi)=\phi\circ\psi\circ\phi^{-1}$ where $\psi \in \operatorname{Aut} (G)$. One can then go to easily show that $\theta$ is an isomorphism.

My question is, is there some reason why considering the map $\theta$ is the right thing to do? How would have I thought to consider that map?

 

[fresh_42]

I have one question that's slightly related to this one, and one which I don't want to create a new thread for. In trying to show that $G \cong H \implies \operatorname{Aut} (G) \cong \operatorname{Aut} (H)$, one supposes that $\phi : G \to H$ and considers the map $\theta:\text{Aut}(G)\rightarrow\text{Aut}(H)$ defined as $\theta(\psi)=\phi\circ\psi\circ\phi^{-1}$ where $\psi \in \operatorname{Aut} (G)$. One can then go to easily show that $\theta$ is an isomorphism.

My question is, is there some reason why considering the map $\theta$ is the right thing to do? How would have I thought to consider that map?

The most naive way is to define $\phi \circ \psi$ or $\psi \circ \phi^{-1}$ but then we land in the wrong group and we have to come back somehow. So the conjugation appears because of this.

Another way is to consider a commutative diagram with inner automorphisms
\begin{aligned}
G &\;\quad \stackrel{\phi}{\longrightarrow} &H\\
\downarrow{\iota(g)}&&\downarrow{\iota(h)}\\
\operatorname{Inn}(G) &\;\quad \stackrel{\theta}{\longrightarrow} &\operatorname{Inn}(H)
\end{aligned}
and make it commute. If there is a formula, substitute the inner by any automorphism.

The third way was my first thought: directly think of conjugation. It is a standard method in group theory, because it solves the problem of coming back. You can automatically think of conjugation whenever the word automorphism is used, either as a construction method like above, or as an example of an automorphism. If something should work for all automorphisms, it has to work for the inner, too.