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Suppose we already know that for any groups $G$ and $H$, that if $f: G \rightarrow H$ is a homomorphism, then if $\langle g \rangle = G$ so too must $\langle f(g)\rangle = H$. Then we would know that if $f$ is a homomorphism, $f$ must "map generators to generators".

My question is to now flip this around and ask, "if we map generators to generators" then are we guaranteed that "$f$ is a homomorphism"?

So letting $G,H$ be groups suppose that $f: G \rightarrow H$ along with the stipulation that if $g \in G$ generates $G$ implies that $f(g)$ generates $H$.

Then we have for any $a,b \in G$ that $a = g^k$ and $b = g^j$ for some $k,j \in \mathbb{Z}$ so that

$$ f(ab) = f(g^k g^j) = f(g^{j+k}) $$

and from here it's not clear to me what we can conclude (other than that $f$ will be surjective). Specifically, I'm curious to know whether we can conclude $f$ is a homomorphism.

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You can use $\langle g\rangle$ for $\langle g\rangle$. – martini Nov 26 '12 at 7:41
Thanks -- just modified question accordingly. – user1770201 Nov 26 '12 at 7:46
up vote 2 down vote accepted

$\def\gen#1{\left\langle#1\right\rangle}\def\Z{\mathbb Z}$No, we can't conclude that $f$ is a homomorphism, as the condition $$\gen g = G \iff \gen{f(g)}= H$$ doesn't tell as anything on $f$'s behaviour on the non-generators. For example let $G = H = \Z/4\Z$, define $f\colon G \to H$ by $f(a+4\Z) = a+4\Z$ for $a \in \{1,3\}$, and $f(a + 4\Z) = 4\Z$ for $a \in \{0,2\}$. Then $f$ maps generators to generators, but it isn't a homomorphism.

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