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I know that for finite groups $G$ and $H$ and $f: G \to H$ homomorphism, for every $x \in G$:

$$ o(f(x))\mid o(x), $$

where $o(x)$ is the order of $x$.

If I consider $G,H$ cyclic, such as $G = {\mathbb Z_{30}}$ and $H = {\mathbb Z}_{50}$, further knowing that $1$ is a generator of ${\mathbb Z}_{30}$ and that it therefore must satisfy that $f(1)$ is generator of ${\mathbb Z}_{50}$, I get that

$$ |H| = o(f(1)) \mid o(1) = |G|, $$

ie. $50 \mid 30$.

What's wrong? Either the deduction is wrong, or the result is to be understood in a special way.

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    $f(1)$ cannot be a generator of $\mathbb{Z}_{50}$ for exactly this reason. The image of a set of generators is a set of generators if and only if $f$ is surjective. There is no surjective homomorphism from $\mathbb{Z}_{30}$ to $\mathbb{Z}_{50}$ – 2017-01-28

2 Answers 2

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$f(1)$ cannot be a generator of $\mathbb{Z}_{50}$ for exactly this reason. The image of a set of generators is a set of generators if and only if $f$ is surjective. There is no surjective homomorphism from $\mathbb{Z}_{30}$ to $\mathbb{Z}_{50}$.

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A homomorphism is determined by the image of the generator of $G$.

Since $o(f(1_G))|o(1_G)=30$, we must have that $f(1_G)$ is mapped to an element of order 1, 2, 3, 5, 10 (these are the divisors of 30 such that there exist elements of such order in $H$.

Note: the number of distinct group homomorphisms between two cyclic groups of order $n$ and $m$ depends on $gcd(m,n)$.