# Find groups $G$, $H$ and a surjective homomorphism $\alpha: G\to H$ such that $\alpha(Z(G)) \neq Z(H)$

Question:

Find groups $G$ and $H$ and a surjective homomorphism $\alpha: G \to H$ such that $\alpha(Z(G)) \neq Z(H)$

Let $G$ and $H$ both be cyclic groups of order 4.

Define $\alpha: G \to H$ such that $g \in G \to I_h \in H$.

So now the center of $G$, $Z(G)$ is all of $G$ and it is getting mapped by $\alpha$ to $I_h$. $Z(H) =$ all of $H = \{I_h, a, a^2, a^3\}$ which is not equal to $\{I_h\}$.

Does that look correct?

* EDIT: Revised answer, does this look correct? *

Let $G$ be the dihedral group or order 8:

$\{I_G, a, a^2, a^3, x, ax, a^2x, a^3x \}$

Let $H$ by the cyclic group of order $8$.

$\{I_G, b, b^2, b^3, b^4, b^5, b^6, b^7 \}$

Define a surjective homomorphism $\alpha:G \to H$ as

$I_G \to I_H$

$a\to b$

$a^2\to b^2$

$a^3\to b^3$

$x\to b^4$

$ax\to b^5$

$a^2x\to b^6$

$a^3x\to b^7$

Now $Z(G) = \{I_G, a^2\}$ and $Z(H) = H$.

Therefore, $\alpha(Z(G)) = \{I_H, b^2\} \neq H = Z(H)$

Does that look correct?

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That's a long way from being surjective. – Chris Eagle Nov 13 '12 at 19:35
Your map $\alpha$ is not surjective. Also, $I_h$ is not a sensible notation for the identity of $H$. You could use $I_H$, though the standard would be $e_H$ or $\text{id}_H$. Finally, as a hint, you should be able to prove that if $G$ and $H$ are abelian and $\alpha$ is surjective, then $\alpha(Z(G)) = Z(H)$ (immediate consequence of the defintions). So your counterexample will have to come from outside the realm of abelian groups. – Michael Joyce Nov 13 '12 at 19:35
I don't think your approach is right. Because the groups you took are both finite and abelian and so every surjective homomorphism is necessarily injective and then your desire result in the title would not be concluded. – S. Snape Nov 13 '12 at 19:37
@sonicboom There are no surjective homomorphisms between these two groups. If there were, such a map would also be injective (since the groups are the same order), and thus an isomorphism. But these groups are not isomorphic. – Brett Frankel Nov 13 '12 at 20:40
$\alpha$ is surjective but it is not a homomorphism. – Martino Nov 14 '12 at 9:02

There are plenty of examples:

• The determinant map $\text{GL}_n(\Bbb R)\rightarrow{\Bbb R}^\times$ when $n$ is even,
• The sign of the permutation map $S_n\rightarrow\{\pm 1\}$
• The orientation map $D_n\rightarrow\{\pm 1\}$ which to a plane isometry $\phi$ in the dihedral group $D_n$ assigns $1$ if and only if $\phi$ preserves orientation.

You should really work out these examples and convince yourself that they satisfy the condition requested.

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No, your homomorphism is not surjective, and so your answer is not correct. In fact, every group of order 4 is abelian, so any surjective homomorphism between groups of order 4 will take the center (the whole group) to the center (the whole group).

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As other people have already pointed out you're not going to get there with abelian groups. But hopefully you can get there with nonabelian groups. You might want to think about groups that don't have a lot of normal subgroups. In particular I would suggest looking at $S_5$, the symmetric group on $5$ elements.

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Hint: The free group on at least $2$ letters has trivial center.

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The easiest example is probably $\alpha: S_3\times C_2\rightarrow C_2$, where you kill the subgroup $A_3\times C_2$.

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Why not just $S_3 \to C_2$? – Derek Holt Nov 13 '12 at 21:39
Sorry, I was thinking of the harder problem to show the center is not a fully invariant subgroup (to which this is the smallest counterexample I think), and is finished by letting $C_2=(1,2)\in S_3$. – user641 Nov 13 '12 at 23:50
$S_n$ with $n \neq 2$ has no center so I don't see how this can be valid? – sonicboom Nov 15 '12 at 2:29
@sonicboom: DO you mean in my answer? Never did I use the group $S_n$ on its own. – user641 Nov 15 '12 at 19:25