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From Wikipedia: An invertible endomorphism of $X$ is called an automorphism. The set of all automorphisms is a subset of $\mathrm{End}(X)$ with a group structure, called the automorphism group of $X$ and denoted $\mathrm{Aut}(X)$. In the following diagram, the arrows denote implication:

enter image description here

Can we give some examples using the integer $\mathbb{Z}$ group (with a closed additive structure, the inverse, the identity 0, and the associative; and also commutative as an abelian group) which satisfy some of the above ---

Please fulfill or correct the following if I am wrong:

  1. The map $\mathbb{Z} \mapsto \mathbb{Z}/2\mathbb{Z}$ (via $k \in \mathbb{Z}$ maps $k \mod 2 \in \mathbb{Z}/2\mathbb{Z}$) is a homomorphism, but not others (not isomorphism, not endomorphism, not automorphism).

  2. The map $\mathbb{Z} \mapsto -\mathbb{Z}$ (via $k \in \mathbb{Z}$ maps to $-k \in \mathbb{Z}$) is a endomorphism and also isomorphism (thus also homomorphism), but not automorphism.

$$\color{red}{\text{But $k \in \mathbb{Z}$ maps to $-k \in \mathbb{Z}$ is invertible, so is it also automorphism?}}$$

  1. The map $\mathbb{Z} \mapsto 2 \mathbb{Z}$ (via $k \in \mathbb{Z}$ maps to $2 k \in \mathbb{Z}$) is an isomorphism (thus also homomorphism), but not endomorphism nor automorphism. Am I correct?

  2. The map $\mathbb{Z} \mapsto \mathbb{Z}$ (via $k \in \mathbb{Z}$ maps to $k \in \mathbb{Z}$) is an automorphism (thus also endomorphism and also isomorphism, homomorphism). Am I correct?

Last Question:

  • Are there examples of homomorphism maps within $\mathbb{Z}$ to itself or subgroup such that it is endomorphism but not isomorphism?

p.s. The automorphism of the group $\mathbb{Z}$ is Aut = $\mathbb{Z}$/2$\mathbb{Z}$, I believe.

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    $\begingroup$ Please ask one question at a time. $\endgroup$
    – Shaun
    May 6 at 21:27
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    $\begingroup$ $-\mathbb{Z}$ and $\mathbb{Z}$ are exactly the same thing. So number $2$ is an automorphism. Number $3$ is an endomorphism which is not an automorphism, because $2\mathbb{Z}$ is a subgroup of $\mathbb{Z}$. $\endgroup$
    – Mark
    May 6 at 21:27
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    $\begingroup$ @annie marie heart Because number $3$ is invertible as a map $\mathbb{Z}\to 2\mathbb{Z}$, not as a map $\mathbb{Z}\to\mathbb{Z}$. It is indeed an isomorphism between $\mathbb{Z}$ and $2\mathbb{Z}$, but it is not an automorphism. $\endgroup$
    – Mark
    May 6 at 22:01
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    $\begingroup$ An endomorphism which is also an isomorphism (when considered with the same codomain) is automatically an automorphism. That's the definition of an automorphism. $\endgroup$
    – Arthur
    May 6 at 22:15
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    $\begingroup$ For number 3, the map ℤ→2ℤ (with a codomain ℤ) is only injective but not surjective; thus not bijective. But for isomorphism, we need to have a bijective homomorphism. So ℤ→2ℤ is NOT isomorphism? (especially to @Torsten Schoeneberg) $\endgroup$ May 6 at 22:36
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Some or all of your questions were answered by our astute commentors.

Yes there is one endomorphism of $\mathbb Z $ which is not an isomorphism, but it's the trivial one. Otherwise, depending on where we send a generator, $\pm1$, we get an endomorphism and an isomorphism. (Recall $\mathbb Z $ is cyclic, and homomorphisms on cyclic groups are determined by where you send a generator. ) Thus we see that for any $n\ne0$, we have $\mathbb Z\cong n\mathbb Z$. If, and only if, we send a generator to a generator, we get an automorphism. Thus there are only two automorphisms. So $\rm {Aut}(\mathbb Z)\cong\mathbb Z_2$.

(Mind you we are talking about $\mathbb Z $ as a group here, not as a ring. That's a whole different discussion. A quite interesting one at that: $\mathbb Z $ is an initial object in the category $\bf {Ring} $ of rings, meaning our hand is forced and there's only one homomorphism from $\mathbb Z $ to $\mathcal R$, for any other ring (with unit). You'll pardon this diversion into Category Theory but, if we relax to the categories of semirings, $\bf {Rig} $, or pseudorings, $\bf {Rng} $ , then, analogous to the situation in $\bf {Grp} $, we no longer have an initial object.)

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  • $\begingroup$ +1. Thanks so much, I posted an answer to see whether people also agree. $\endgroup$ May 7 at 4:26
  • $\begingroup$ Is there any nontrivial map example such that for (1) homomorphism, (2) isomorphism, (3) endomorphism, and (4) automorphism, we get (1) O, (2) O, (3) X, (4) X? (see below) $\endgroup$ May 7 at 4:26
  • $\begingroup$ In your setup everything is an endomorphism, that's just a map from the space back into itself. $\endgroup$
    – user403337
    May 7 at 4:43
  • $\begingroup$ Can you give another set up (not my ℤ↦ℤ) such that for (1) homomorphism, (2) isomorphism, (3) endomorphism, and (4) automorphism, we get (1) O, (2) O, (3) X, (4) X? (see below) $\endgroup$ May 7 at 4:53
  • $\begingroup$ What about an isomorphism $\mathbb Z\cong2\mathbb Z $? $\endgroup$
    – user403337
    May 7 at 4:57
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Below in our examples, we consider the group homomorphism between $\mathbb{Z} \mapsto \mathbb{Z} $ where the first $k \in \mathbb{Z}$ forms the domain, the second $\mathbb{Z}$ is the codomain, and $f(k)$ is the image.

We ask whether it is (1) homomorphism, (2) isomorphism, (3) endomorphism, and (4) automorphism

We will give "O" if it is true. We will give "X" if it is false.

  1. The domain $k \in \mathbb{Z}$ maps to the image $f(k)=0$. It is not injective nor surjective over codomain.

$$\text{(1) O, (2) X, (3) O, (4) X.}$$

  1. The domain $k \in \mathbb{Z}$ maps to the image $f(k)=k \mod N \in \mathbb{Z}/N\mathbb{Z}$, where $N$ can be some integer. In fact, the image $\mathbb{Z}/N\mathbb{Z}$ is not a subgroup of codomain. So we cannot consider endomorphism. It is not injective but it is surjective over image.

$$\text{(1) O, (2) X, (3) X, (4) X.}$$

  1. The domain $k \in \mathbb{Z}$ maps to the image $f(k)= N k\in N\mathbb{Z}$, where $N$ can be some integer but $N \neq \pm 1$. In fact, the image $ N\mathbb{Z}$ is a subgroup of codomain. So it is an endomorphism. It is injective and also surjective over image. It is injective but not surjective over the codomain.

$$\text{(1) O, (2) O, (3) O, (4) X.}$$

  1. The domain $k \in \mathbb{Z}$ maps to the image $f(k)= \pm k\in \mathbb{Z}$. In fact, the image $\mathbb{Z}$ is the codomain. It is injective and also surjective over image. It is injective and surjective over the codomain.

$$\text{(1) O, (2) O, (3) O, (4) O.}$$

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