The inclusion map not always a group homomorphism $i: \mathbb{Z}/3\mathbb{Z} \to\mathbb{Z}/6\mathbb{Z}$ Consider the inclusion map $i: \mathbb{Z}/3\mathbb{Z} \to\mathbb{Z}/6\mathbb{Z}, \, a \mapsto a$, Question: is this a group homomorphism?
$i$ certainly satisfies $i(a+b) = a + b = i(a) + i(b)$ but $i$ is not a well defined map: Since $\bar{0} = \bar{3}$, $i(\bar{0}) = \bar{0}$ but $i(\bar{3}) = \bar{3} \neq \bar{0}$.
So $i: \mathbb{Z}/3\mathbb{Z} \to\mathbb{Z}/6\mathbb{Z}$, the inclusion map is not a group homomorphism in this case; this seemed surprising to me at first, does this look correct or am I missing something?
Edit: Instead of "inclusion" and "map", I was thinking "embedding" and "set map".
 A: The correct way to think about homomorphisms $\newcommand\Z{\mathbb Z}\Z/3\Z\to G$ is to first look at a homomorphism $f\colon \Z\to G$. Then, by the fundamental homomorphism theorem there exists a group homomorphism $\bar f\colon \Z/3\Z\to  G$ given by $\bar f(a + 3\Z) = f(a)$ if and only if $3\Z \subseteq \ker f$. As the comments suggest, the problem is not that $\bar f$ could ever fail to be homomorphism, it can fail to be a map in the first place since you need to have $a + 3\Z = b + 3\Z \implies \bar f(a + 3\Z) = \bar f(b + 3\Z)$ which boils down precisely to the condition $3\Z\subseteq \ker f$.
You can convince yourself easily that all group homomorphisms $g\colon \Z/3\Z \to G$ are of this form just by considering $f\colon \Z \to G$ given by $f(a) = g(a+ 3\Z)$ and noting that $\bar f = g$.
Now, for your concrete example, you want to first consider a group homomorphism $f\colon \Z \to \Z/6\Z$ given by the canonical projection $f(a) = a + 6\Z$. Note that defining $f(a) = a$ is simply wrong since elements of $\Z$ and $\Z/6\Z$ are completely different things, even though we do sometimes abuse this notation, but then one needs to keep track of it mentally not to make mistakes. Kernel of $f$ is simply $6\Z$, so in order to $\bar f$ (given by $\bar f(a + 3\Z) = a + 6\Z$) be a well-defined map, we need to have $3\Z\subseteq 6\Z$, i.e. $3\mid x \implies 6\mid x$, which is not true. Therefore, $\bar f$ is not even a map.
To demonstrate this more concretely, note that $0 + 3\Z = 3 + 3\Z$ since $3\in 3\Z$. However, $$\bar f(0 + 3\Z) = f(0) = 0 + 6\Z \neq 3 + 6\Z = f(3) = \bar f(3+3\Z),$$ so $\bar f$ is not a function at all. We say that it's not well-defined since we tried to define it, but couldn't make it work.

Interestingly, there is a way to think of $\Z/3\Z$ as subgroup of $\Z/6\Z$ by considering $f\colon \Z\to \Z/6\Z$ given by $f(a) = 2a + 6\Z$. You can see that $\ker f = 3\Z$, so $\bar f(a + 3\Z) = 2a + 6\Z$ is not just a well-defined group homomorphism, it is a monomorphism, again by the fundamental homomorphism theorem.
A: From Wikipedia, if $A$ is a subset of $B$, then the inclusion map is defined by $\iota:A\rightarrow B$ where $\iota(x)=x$ for each $x\in A$. Therefore we can define inclusion map only when $A$ is a subset of $B$.
Let $n$ be a positive integer. Note that $$\Bbb{Z} / n\Bbb{Z}=\{\bar{0},\bar{1},\dots, \overline{n-1}\}$$
where for each $\bar{g}\in \Bbb{Z} / n\Bbb{Z}$, $\bar{g}=\{g+nz:z\in \Bbb{Z}\}$.
In your question, $\Bbb{Z/3\Bbb{Z}}$ is not a subset of $\Bbb{Z/6\Bbb{Z}}$. In fact, all elements of $\Bbb{Z/3\Bbb{Z}}$ are not elements of $\Bbb{Z/6\Bbb{Z}}$. For example, $\bar{0}$ in $\Bbb{Z/3\Bbb{Z}}$ consists of all multiples of $3$, while in $\Bbb{Z/6\Bbb{Z}}$, $\bar{0}$ consists of all multiples of $6$. This means that $\bar{0}$ in $\Bbb{Z/3\Bbb{Z}}$ is not the same as $\bar{0}$ in $\Bbb{Z/6\Bbb{Z}}$.
