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