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Let $G$ and $K$ be (possibly non-Abelian) groups and let $\phi:G\rightarrow K$ be a homomorphism. Let $\bar{G}$ be a group containing $G$ as a subgroup.

Is it always possible to extend $\phi$ to a homomorphism $\bar{\phi}:\bar{G}\rightarrow K$ (such that $\bar{\phi}$ restricted to $G$ is $\phi$ itself)? If not what conditions are required for this to be true?

I have seen somewhere that for the Abelian case if $K$ is divisible this is possible but I'm mostly interested in the non-Abelian case.

Is it always possible to extend $\phi$ to a homomorphism $\bar{\phi}:\bar{G}\rightarrow \bar{K}$ where $\bar{K}$ is some group containing $K$ as a subgroup? If not what conditions are required for this to be true?

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up vote 6 down vote accepted

The answer to your first question is "no", even if you assume $G$, $\bar{G}$, and $K$ are all isomorphic! Take for example $\bar{G} = K = \mathbb{Z}$, $G \subset \bar{G}$ to be the subgroup generated by $\{2\}$, and let $\phi$ map $2 \in \bar{G}$ to $1 \in K$.

If you're allowed to extend $K$, then you can always find $\bar{\phi}$ by the following construction: Let $A$ be a generating set of $G$, and let $\bar{A}$ be such that the set $A \cup \bar{A}$ generates $\bar{G}$. Now take an explicit presentation of $\bar{G}$ with this generating set, and call the set of relations $R$.

Next, let $B \subset K$ be such that the set $B \cup \phi(A)$ generates $K$, and take an explicit presentation of $K$ using this generating set. Call the set of relations $R'$.

To construct $\bar{K}$, we'll form a new presentation. The generating set will be (symbolically) $A \cup \bar{A} \cup B$, and the relations will be everything in $R'$ (with the symbol $\phi(a)$ replaced by the symbol $a$), together with everything in $R$. Then $\bar{K}$ clearly contains $K$ as a subgroup (it's the subgroup generated by $A \cup B$).

Let $\bar{\phi}: \bar{G} \rightarrow \bar{K}$ then be the homomorphism defined to take generators in $\bar{G}$ to their counterparts in $\bar{K}$. This is indeed a homomorphism since relations in $\bar{G}$ are satisfied by their images, and it agrees with $\phi$ on $G$.

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No. An easy counterexample: consider the dihedral group $D_4 = \langle r,s \mid r^4 = s^2 = 1, \: rs = sr^{-1} \rangle$. Consider a morphism $\chi \colon D_4 \to \mathbb C^*$; then $\chi(s) \in \{\pm 1\}$, while $\chi(r) \in \{\pm 1, \pm i\}$. Set $w = \chi(r)$, $z = \chi(s)$. Then $wz = \chi(rs) = \chi(sr^3) = zw^3$, hence $w^2 = 1$, i.e. $\chi(r) \in \{\pm 1\}$.

Now, consider $H = <r> \subset D_4$ and let $\varphi \colon H \to \mathbb C^*$ be defined by $\varphi(r) := i$. This is well defined, and it cannot be extended.

By the way, this counterexample has the merit that the landing group is $\mathbb C^*$, which is divisible. In an abelian case, every morphism toward $\mathbb C^*$ can be extended, because over a PID divisible implies injective (thanks to Baer's criterion).

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Thanks, your comment was helpful. – Jim Dec 4 '12 at 23:43
I encounter a similar but seemingly different question: Given a homomorphism $\phi: G\to K$ where $K$ is Abelian, how to get a new homomorphism $\varphi:\bar{G}\to\bar{K}$, such that: (1) $G$ and $K$ can be embedded into $\bar{G}$ and $\bar{K}$, respectively; (2) $\varphi(bar{G})$ is a non-Abelian subgroup of $\bar{K}$. – Pigmann Mar 1 '14 at 11:27

$A_4$ has a normal subgroup of $4$ elements, hence, a homomorphism onto $C_3$. $A_4$ is a subgroup of $S_4$. $S_4$ has no normal subgroup of order $8$, hence, no homomorphism to $C_3$ extending the one defined on $A_4$.

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That's a very easy example, thanks! – Jim Dec 4 '12 at 23:42

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