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

• Mr Davis 97
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.

## Homework Statement

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

## 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)##

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
fresh_42 said:
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?

Last edited:
Mr Davis 97 said:
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}