Isomorphic image of a centre of a group

Let $G$ and $G'$ be groups. Let $\phi: G \to G'$ be an isomorphism. Is it true that $\phi(Z(G))= Z(\phi(G))$.

Let $x \in Z(G)$. Then for any $g\in G$ we have $\phi(x)\phi(g) = \phi(xg) = \phi(gx) = \phi(g)\phi(x)$. Hence $\phi(Z(G)) \subseteq Z(\phi(G))$

Let $x \in Z(\phi(G))$. Then $x = \phi(a)$ for some $a\in G$. For any $\phi(g) \in \phi(G)$, we have $\phi(a)\phi(g) = \phi(g)\phi(a)$. Thus $\phi(ag) = \phi(ga)$. As $\phi$ is injective $ag =ga$. Hence $a \in Z(G)$. Thus $x = \phi(a) \in \phi(Z(G))$.

Is this proof correct? I'm not sure about the second inclusion

• Why not apply your first argument to the inverse map $\phi^{-1}$? – Lord Shark the Unknown Jun 9 '17 at 5:59
• @LordSharktheUnknown I think that is a very good way to prove the statement, but I think that it is instructive (besides what the OP wants) to see how to make this style of proof work. – Alex Ortiz Jun 9 '17 at 6:12

I think this preserves the style of proof you are going for and clears things up:

Let $x\in Z(G')$. Note that $x = \phi(a)$ for some $a\in G$. We claim that $a$ is an element of the center of $G$. To that end, let $b\in G$ be arbitrary. Then $\phi(b)\in G'$ so $\phi(ab) = \phi(a)\phi(b) = \phi(b)\phi(a) = \phi(ba)$ by our hypothesis on $x = \phi(a)$. Now $\phi$ is an injection, so $ab = ba$. Since $b$ was arbitrary, $a\in Z(G)$, as desired.

Yes. Centres of isomorphic groups are isomorphic.

Intuitively this is because if an element in $G$ commutes with every other element then its image under an isomorphism commutes with everything in the image of $G,$ and similarly if we start with an element in the image of $G,$ then we apply the same argument again (but swap the isomorphism for its inverse - doesn't matter, it's essentially the same isomorphism).

It is quite easy to translate from plain language into a formal proof.

Forgot that you asked two questions:

You already showed that $\phi(Z(G))\subset Z(\phi(G)).$ Now you can swap the groups around and repeat the same argument with $G\leftrightarrow\phi(G)$ and $\phi\leftrightarrow\phi^{-1}.$ Your argument makes the same statement in principle anyway, and shows that $Z(\phi(G))\subset\phi(Z(G)).$ Then you can say $\phi$ is an isomorphism so $Z(\phi(G))\cong\phi(Z(G)).$