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I have a question about group endomorphism. Suppose $G$ is a group and endomorphism $f : G \to G$. Is it true that $f(Z(G)) = Z(G)$ ?

If not, suppose I have $G= D_{8}$ dihedral group of order $8$. Can we find an endomorphism of $D_{8}$ that maps $Z(D_{8}) \to D_{8} \setminus Z(D_{8})$ ?

Thanks in advance.

Shaun
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NightShadow
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2 Answers2

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Neither of those claims are true. For the first one, consider the trivial homomorphism. For the second, the image of a homomorphism can't possibly not contain the identity, as a homomorphism always maps the identity to the identity; and the center always contains the identity.

Alex Provost
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  • Okey the first statement is clear. For the second statement, i do understand that homomorphism maps the identity to the identity. But i mean for $D_8$, we have $Z(D_8) = {e,a^2}$ where a is rotation 90 degree. Can we find $f$ as endomorphism from $D_8$ to $D_8$ that maps $a^2$ to $D_8 - {e,a^2}$? – NightShadow Nov 30 '20 at 05:23
  • @JerikoGormantara Asking whether a single element in the center is still mapped to the center is not the same thing as asking whether the whole center can be mapped outside the center as in your original question. For $D_8$, if your $f(a)$ has order 2 then $f(a^{2}) = 1$, and if your $f(a)$ has order 4 then either $f(a) = a$ or $f(a) = a^3$, and then $f(a^2) = a^2$; in all cases $f(a^2) \in {1,a^2}$. – Alex Provost Nov 30 '20 at 05:43
  • Ah yes pardon me for that, Sir. Right, thank you for the expalanation. – NightShadow Nov 30 '20 at 07:09
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Take the map $f:D_8 \rightarrow D_8/\langle r\rangle \cong \mathbb{Z/2Z}$.

$Z(D_8) = \langle r^2 \rangle$ but $f(Z(G))=\{0\} \neq \mathbb{Z/2Z} =Z(\mathbb{Z/2Z})$ because $\mathbb{Z/2Z}$ is abelian.

We know that the subgroup generated by $r^2$ is of order $2$. Hence we have a map $g:\mathbb{Z/2Z} \cong \langle r^2 \rangle \rightarrow D_8 $ which is the inclusion map. So $g\circ f:D_8 \rightarrow D_8$ serves as a counter-example to the first question $f(Z(G))=Z(G)$

As for the second question whether there is an endomorphism $f:D_8 \rightarrow D_8$ such that it maps $Z(D_8)$ to some element in $D_8-Z(D_8)$? This cannot happen.

The centre of $D_8$ is $\{1,r^2\}$. Any endomorphism of $D_8$ can map $r$ to an order $4$ element or order $2$ element. Only order $4$ element in $D_8$ are $r,r^3$. Hence if $r$ is mapped to either of them we can see that $r^2$ is mapped to itself. When $r$ is mapped to any order $2$ element then $r^2$ is mapped to identity element of $D_8$ which is again in the centre. Hence in both cases $f(Z(D_8)) \subset Z(D_8)$

nrynn
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