Do Isomorphic Groups Have Isomorphic Centers?

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In summary: 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.
  • #1
Mr Davis 97
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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)##
 
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  • #2
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".
 
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  • #3
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?
 
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  • #4
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}
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.
 
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Related to Do Isomorphic Groups Have Isomorphic Centers?

1. What does it mean for two groups to be isomorphic?

Groups G and H are isomorphic if there exists a bijective homomorphism (a mapping that preserves the group operation) from G to H. This means that the two groups have the same structure and elements, just labeled differently.

2. How do you prove that two groups are isomorphic?

To prove that two groups G and H are isomorphic, you must show that there exists a bijective homomorphism from G to H. This can be done by constructing a mapping that preserves the group operation and showing that it is both one-to-one and onto.

3. What does it mean for the centers of two groups to be isomorphic?

If G and H are two groups with isomorphic centers, Z(G) ~ Z(H), this means that the centralizers (subgroups of elements that commute with all elements of the group) of each group are isomorphic. In other words, the elements that commute with all elements of G are the same as the elements that commute with all elements of H.

4. How do you prove that the centers of two groups are isomorphic?

To prove that the centers of two groups G and H are isomorphic, you must show that there exists a bijective homomorphism from Z(G) to Z(H). This can be done by constructing a mapping that preserves the group operation and showing that it is both one-to-one and onto.

5. What is the significance of two groups having isomorphic centers?

If G and H have isomorphic centers, this means that the centralizers of each group are the same. This can provide helpful information when studying the structure and properties of the groups, as the centralizers play an important role in understanding the behavior of elements within a group.

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