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There are homomorphisms of a group into itself that are one-to-one but not onto. Example :

$f:\Bbb Z \rightarrow \Bbb Z$ defined by $f(n)=2n$. $f$ is a homomorphism with kernel ${0}$. Therefore it is one-to-one. Image of the homomorphism is $2\Bbb Z$. Therefore it is not onto.

Is there an example of a homomorphism of a group onto itself that is not one-to-one?

It has to be an infinite group $G$ having a normal subgroup $N$ such that $G/N≈G$. Index of this normal subgroup $N$ in $G$ must be infinite.

Not Euler
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7 Answers7

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Try these:

  • $\mathbb C^\times \to \mathbb C^\times$ given by $z \mapsto z^2$

  • $\mathbb R[x] \to \mathbb R[x]$ given by $f \mapsto f'$

lhf
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Let $G$ be the quotient group $G = \mathbb{Q} / \mathbb{Z}$.

For any nonzero $n$, we can define $f : G \to G$ to be the multiplication-by-$n$ homomorphism. This will be surjective whenever $n \neq 0$, but it is an isomorphism only when $n = \pm 1$.

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For the direct sum $\oplus_{i = 0}^{\infty} \Bbb Z$ of integer sequences with finitely many nonzero entries (under entrywise addition), the left-shift homomorphism, $$(a_0, a_1, a_2, \ldots) \mapsto (a_1, a_2, a_3, \ldots) ,$$ is onto but not one-to-one.

(Dually, the right-shift homomorphism, $$(a_0, a_1, a_2, \ldots) \mapsto (0, a_0, a_1, \ldots) ,$$ is one-to-one but not onto.)

Travis Willse
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Consider the map from the unit circle to itself sending $\theta \mapsto 2\theta$.

John Hughes
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We can take familiar set-theroetic examples of the analogous phenomenon and transport them over to groups.

Let $X$ be an infinite set and $g : X \to X$ be any surjective map that is not invertible.

Then, you can take $G = F_X$ to be the free group generated by $X$ and $f$ to be the map $F_X \to F_X$ induced by $g$ acting on generators.

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Another interesting example is the Prüfer $p$-group $G$: it is the abelian group with countably infinite generators $p_1,p_2,\ldots,$ satisfying $p_1^p = 1$ and $p_{i+1}^p = p_i$ for all $i>1$. (In fact these relations imply $G$ is abelian.) The map $x \mapsto x^p$ is a surjective group homomorphism (since all generators are in its image), but is not injective since $p_1$ is mapped to $1$.

Joshua Mundinger
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  • This is (almost) the fundamental group of the infinite mapping cylinder of $f: S^1 \to S^1: z \mapsto z^p$; [that group has an extra $\Bbb Z$ at the "finite" end]. The map described above corresponds to translating the cylinder one unit towards its finite end (with a little fiddling to deal with what happens at the finite end). In that sense, it's very similar to @Travis's example, in the sense that both are induced by a left-shift (roughly). – John Hughes Jun 13 '18 at 15:49
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Similar to lhf's example, if $G$ is a (nontrivial) group, and we endow $\mathbb{G}:=G^\mathbb{N}$ with componentwise multiplication, then the shift map $\sigma:\mathbb{G}\to\mathbb{G}$ sending $(g_n)_{n\in\mathbb{N}}$ to $(g_{n+1})_{n\in\mathbb{N}}$ is onto and not one to one.

Olivier Bégassat
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