So far we've studied groups one at a time. This page is about the maps between groups that respect the group structure, and they turn out to be the best tool we have for comparing two groups. A note on conventions before we start: as on Symmetric Groups, products of permutations are composed right to left, so in $\sigma\tau$ we apply $\tau$ first and then $\sigma$.
A homomorphism is a function that doesn't care whether you multiply before or after you apply it.
Homo-morph-ism. "Same-shape-ism." Someone looked at "structure-preserving map" and decided it needed more Greek.
Suppose $(G, \circ)$ and $(H, *)$ are groups. A map $\phi : G \rightarrow H$ which satisfies
$$\phi(x \circ y) = \phi(x) * \phi(y), \quad \forall x, y \in G,$$
is called a homomorphism from $G$ to $H$.
Read the equation from left to right: on the left we multiply in $G$ (using $\circ$) and then apply $\phi$; on the right we apply $\phi$ first and then multiply in $H$ (using $*$). A homomorphism is a map where both routes give the same answer. Once we know a map is a homomorphism, the usual words for functions give it new names.
Injective, surjective, and bijective all come from Functions. Nothing new about them here: we're just giving homomorphisms with those properties fancier names.
A homomorphism $\phi : G \rightarrow H$ is called a monomorphism if it is injective.
A monomorphism copies $G$ faithfully into $H$, with no two elements landing on top of each other. Flip the condition around and you get:
A homomorphism $\phi : G \rightarrow H$ is called an epimorphism if it is surjective.
When a homomorphism is both, the two groups are really the same group with the elements renamed.
On Groups of Order 1–3, "isomorphic" meant renaming the elements so that one multiplication table turns into the other. A bijective homomorphism is exactly that renaming, written as a formula.
A homomorphism $\phi : G \rightarrow H$ is called an isomorphism if it is bijective.
Notation: if there is an isomorphism from $G$ to $H$, we write $G \cong H$.
And when a group gets renamed as itself:
An isomorphism $\phi : G \rightarrow G$ is called an automorphism of $G$.
Every group has at least one automorphism (the identity map). Let's look at a pile of examples, starting with the laziest map imaginable.
The trivial homomorphism sends everything to $e$ and still passes every test. Absolute legend of doing the bare minimum.
For any groups $G$ and $H$, show that the map $\phi : G \rightarrow H$ defined by $\phi(g) = e_H$, $\forall g \in G$, is a homomorphism.
$\forall x, y \in G$, $\phi(xy) = e_H = e_H \cdot e_H = \phi(x) \cdot \phi(y)$.
This is called the trivial homomorphism.
The two groups in a homomorphism don't have to use the same operation. Here's a famous one that turns addition into multiplication.
$G = (\mathbb{R}, +)$, $H = (\mathbb{R}_{>0}, \cdot)$ (positive reals). Define $\phi : G \rightarrow H$ by $\phi(x) = e^x$. Is $\phi$ a homomorphism?
$\forall x, y \in \mathbb{R}$,
$$\phi(x+y) = e^{x+y} = e^x e^y = \phi(x) \cdot \phi(y)$$
(using the properties of exponentials, the binary operation in $G$, and the binary operation in $H$).
Also, $\phi$ is a bijection (its inverse is $\log$), so it is an isomorphism.
The rule $e^{x+y} = e^x e^y$ that you learned years ago was a homomorphism all along. The determinant has a famous rule like that too.
$G = (GL_2(\mathbb{R}), \cdot)$, $H = (\mathbb{R} \setminus \{0\}, \cdot)$. Define $\phi : G \rightarrow H$ by $\phi(A) = \det(A)$. Is $\phi$ a homomorphism?
$\forall A, B \in G$,
$$\phi(AB) = \det(AB) = \det(A)\det(B) = \phi(A)\phi(B)$$
(using the properties of $\det$, multiplication of $2 \times 2$ matrices, and multiplication of real numbers).
Here, $\phi$ is an epimorphism, but it is not injective. (Every $t \neq 0$ is $\det\begin{pmatrix} t & 0 \\ 0 & 1 \end{pmatrix}$, but $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ and $\begin{pmatrix} 2 & 0 \\ 0 & 1/2 \end{pmatrix}$ both have determinant $1$.)
The integers give a whole family of homomorphisms, one for each $n$, and they behave differently depending on $n$.
$G = H = (\mathbb{Z}, +)$, $n \in \mathbb{Z}$. Define $\phi : G \rightarrow H$ by $\phi(k) = nk$. Is $\phi$ a homomorphism? Is it injective? Surjective?
$\forall k, l \in \mathbb{Z}$, $\phi(k+l) = n(k+l) = nk + nl = \phi(k) + \phi(l)$. So $\phi$ is a homomorphism:
Reducing mod $n$ is a homomorphism too. This is the reason modular arithmetic works at all.
$G = \mathbb{Z}$, $H = \mathbb{Z}/n\mathbb{Z}$. Define $\phi : G \rightarrow H$ by $\phi(k) = \overline{k}$ (where $\overline{k} = k + n\mathbb{Z}$). Is $\phi$ a homomorphism?
$\forall k, l \in \mathbb{Z}$, $\phi(k+l) = \overline{k+l} = \overline{k} + \overline{l} = \phi(k) + \phi(l)$ (by the definition of $+$ in $H$).
Here, $\phi$ is an epimorphism, but it is not injective ($\phi(0) = \phi(n) = \overline{0}$).
Next, a homomorphism between two groups we already suspected were "the same."
On Subgroups & Lagrange's Theorem we showed $x^i = x^j \iff i \equiv j \pmod n$ when $|x| = n$, and Cyclic Groups used it to prove that any two cyclic groups of the same order are isomorphic. Example 6 is that proof in miniature.
$G = C_n = \langle x \mid x^n = e \rangle$, $H = \mathbb{Z}/n\mathbb{Z}$. Show that $C_n \cong \mathbb{Z}/n\mathbb{Z}$.
Recall: $G = \{e, x, x^2, \dots, x^{n-1}\}$, and $x^i = x^j \iff i \equiv j \pmod n$.
Define $\phi : G \rightarrow H$ by $\phi(x^k) = \overline{k}$. This is well defined: if $x^k = x^l$ then $k \equiv l \pmod n$, so $\overline{k} = \overline{l}$.
Then $\phi(x^k x^l) = \phi(x^{k+l}) = \overline{k+l} = \overline{k} + \overline{l} = \phi(x^k) + \phi(x^l)$.
Also, $\phi$ is a bijection, so $C_n \cong \mathbb{Z}/n\mathbb{Z}$.
(This shows that any two finite cyclic groups of the same order are isomorphic.)
Now multiply inside $\mathbb{Z}/n\mathbb{Z}$ instead of inside $\mathbb{Z}$. Unlike Example 4, injective and surjective now come as a package deal.
$G = H = \mathbb{Z}/n\mathbb{Z}$, $a \in \mathbb{Z}$. Define $\phi : G \rightarrow H$ by $\phi(k) = ak \pmod n$. Is $\phi$ a homomorphism? When is it injective? Surjective?
$\phi(k+l) = a(k+l) = ak + al = \phi(k) + \phi(l)$. Note that $\phi$ is:
Therefore:
Both halves of the surjective bullet are the theorem on solving $ax \equiv b \pmod n$ from Modular Arithmetic: a solution exists exactly when $(a, n)$ divides $b$.
See it for yourself. Each dot on the top row is sent to $ak$ on the bottom row. Watch what happens to the arrows when $a$ shares a factor with $n$.
Change $n$ and $a$. The kernel (everything sent to $0$) and the image (everything that gets hit) light up as you go.
Our last example works in every group, and it's one you'll meet again very soon.
$ghg^{-1}$ is the conjugate of $h$ by $g$. Conjugation is the engine behind Normal Subgroups.
Let $G$ be a group and $\forall g \in G$ define $\tau_g : G \rightarrow G$ by $\tau_g(h) = ghg^{-1}$ (conjugation by $g$). Show that $\tau_g$ is an automorphism of $G$.
$\forall g \in G$, $\tau_g$ is:
Therefore, $\forall g \in G$, $\tau_g$ is an automorphism of $G$.
Not every map is a homomorphism, of course. To check one by brute force, you'd test $\phi(xy) = \phi(x)\phi(y)$ for every pair $(x, y)$, and that's a whole grid of checks. The test below does exactly that.
Pick a map. Square $(x, y)$ turns purple when $\phi(xy) = \phi(x)\phi(y)$ and red when it doesn't. Tap any square to see the computation.
A single red square is enough to kill a homomorphism.
Which of these maps are homomorphisms?
The definition only talks about products. But because identities and inverses are defined in terms of products, a homomorphism automatically respects those too. Throughout this section, suppose $\phi : G \rightarrow H$ is a homomorphism.
Each part leans on the one before it.
Proof of (a). $\forall x \in G$,
$$\phi(x) = \phi(x \cdot e_G) = \phi(x) \cdot \phi(e_G)$$
(the first step by the definition of the identity, the second because $\phi$ is a homomorphism). So $\phi(x) e_H = \phi(x)\phi(e_G)$, and the cancellation law gives $e_H = \phi(e_G)$. $\blacksquare$
Proof of (b). $\phi(g^{-1})\phi(g) = \phi(g^{-1}g) = \phi(e_G) = e_H$ (by (a)) $\Rightarrow \phi(g^{-1}) = \phi(g)^{-1}$. $\blacksquare$
Proof of (c). First suppose $n \ge 0$, and use induction on $n$.
Now suppose $n < 0$, and write $n = -m$ with $m \in \mathbb{N}$. Then, using the case we just proved and (b),
$$\phi(g^{-m}) = \phi\big((g^{-1})^m\big) = \phi(g^{-1})^m = \big(\phi(g)^{-1}\big)^m = \phi(g)^{-m}. \quad \blacksquare$$
The next two properties say that a homomorphism doesn't just preserve elements, it carries subgroups along with it.
$\phi(G) \le H$ (the image of a homomorphism is a subgroup of the codomain).
The three checks below (non-empty, closed under multiplication, closed under inverses) are the subgroup criterion from Subgroups & Lagrange's Theorem.
Proof. Use the subgroup criterion. $\phi(G)$ is:
Therefore $\phi(G) \le H$. $\blacksquare$
$G = H = \mathbb{Z}$, $\phi : G \rightarrow H$, $\phi(k) = nk$ ($n \in \mathbb{Z}$). What subgroup is $\phi(G)$?
$\phi$ is a homomorphism (Example 4) $\Rightarrow \phi(G) = n\mathbb{Z} = \{nk : k \in \mathbb{Z}\} \le \mathbb{Z}$.
The image lives in the codomain. There's a matching subgroup in the domain: everything that gets crushed down to the identity.
The kernel of $\phi$ is $\ker(\phi) = \{g \in G : \phi(g) = e_H\}$.
In Multiplier Rows, the kernel was the set of blue dots on the top row.
$\ker(\phi) \le G$ (the kernel of a homomorphism is a subgroup of the domain).
Proof. Use the subgroup criterion again. $\ker(\phi)$ is:
Therefore $\ker(\phi) \le G$. $\blacksquare$
$G = (GL_2(\mathbb{R}), \cdot)$, $H = (\mathbb{R} \setminus \{0\}, \cdot)$, $\phi : G \rightarrow H$, $\phi(A) = \det(A)$. Find $\ker(\phi)$.
$$\ker(\phi) = \{A \in G : \phi(A) = e_H\} = \{A \in GL_2(\mathbb{R}) : \det(A) = 1\} = SL_2(\mathbb{R}) \le G.$$
Kernels are actually much more special than ordinary subgroups. That's the main story of Normal Subgroups, where it turns out every kernel is a normal subgroup.
Let $\phi : \mathbb{Z}/12\mathbb{Z} \rightarrow \mathbb{Z}/12\mathbb{Z}$, $\phi(k) = 3k \pmod{12}$. Find $\ker(\phi)$ and $\phi(\mathbb{Z}/12\mathbb{Z})$.
$\phi$ is a homomorphism by Example 7. $3k \equiv 0 \pmod{12} \iff k \equiv 0 \pmod 4$, so $\ker(\phi) = \{0, 4, 8\}$.
The image is $\{3k \bmod 12 : k \in \mathbb{Z}/12\mathbb{Z}\} = \{0, 3, 6, 9\}$.
Notice that $|\ker(\phi)| \cdot |\phi(\mathbb{Z}/12\mathbb{Z})| = 3 \cdot 4 = 12$. That's no coincidence, and you can check it on every setting of Multiplier Rows.
Suppose $\phi : G \rightarrow H$ is a homomorphism and $G = \langle g \rangle$ is cyclic. Show that $\phi(G)$ is cyclic.
Proof. Every element of $G$ is $g^n$ for some $n \in \mathbb{Z}$, so every element of $\phi(G)$ is $\phi(g^n) = \phi(g)^n$ (by 1c). Therefore $\phi(G) = \{\phi(g)^n : n \in \mathbb{Z}\} = \langle \phi(g) \rangle$, which is cyclic. $\blacksquare$
One more property: if a homomorphism can be undone, the undo button is a homomorphism too.
If $\phi : G \rightarrow H$ is an isomorphism then its inverse function $\phi^{-1} : H \rightarrow G$ is also an isomorphism.
Proof. $\phi^{-1}$ is a bijection, so we just need to show that it is a homomorphism. Suppose $h_1, h_2 \in H$, $\phi^{-1}(h_1) = g_1$, $\phi^{-1}(h_2) = g_2$. Then $\phi(g_1) = h_1$ and $\phi(g_2) = h_2$ (definition of the inverse function).
$$\Rightarrow h_1 h_2 = \phi(g_1)\phi(g_2) = \phi(g_1 g_2) \quad (\phi \text{ is a homomorphism})$$
$$\Rightarrow \phi^{-1}(h_1 h_2) = g_1 g_2 = \phi^{-1}(h_1)\phi^{-1}(h_2) \quad (\text{definition of the inverse function}). \quad \blacksquare$$
In particular, $G \cong H$ implies $H \cong G$, so "isomorphic" is a symmetric relation, as it should be. Applying Property 4 to Example 2 gives something familiar.
So the log rule I memorised in high school was a group isomorphism the whole time? Nobody told me.
$G = (\mathbb{R}, +)$, $H = (\mathbb{R}_{>0}, \cdot)$, $\phi : G \rightarrow H$, $\phi(x) = e^x$. What does Property 4 say about $\phi^{-1}$?
$\phi$ is an isomorphism $\Rightarrow \phi^{-1} : \mathbb{R}_{>0} \rightarrow \mathbb{R}$ is an isomorphism.
Note: $\phi^{-1}(x) = \log x$, so $\phi^{-1}(xy) = \phi^{-1}(x) + \phi^{-1}(y) \Rightarrow \log(xy) = \log x + \log y$.
Let $G$ be a group and define $\iota : G \rightarrow G$ by $\iota(g) = g^{-1}$. Prove that $\iota$ is an automorphism if and only if $G$ is Abelian.
Proof. $\iota$ is always a bijection, since $\iota(\iota(g)) = (g^{-1})^{-1} = g$ makes it its own inverse. So $\iota$ is an automorphism exactly when it is a homomorphism, i.e. when
$$(gh)^{-1} = g^{-1}h^{-1}, \quad \forall g, h \in G.$$
Since $(gh)^{-1} = h^{-1}g^{-1}$, this says $h^{-1}g^{-1} = g^{-1}h^{-1}$ for all $g, h \in G$. As $g$ and $h$ range over $G$, so do $g^{-1}$ and $h^{-1}$, so this holds if and only if every pair of elements of $G$ commutes, i.e. $G$ is Abelian. $\blacksquare$
Here's the payoff of all that machinery: every group, no matter how abstract it looks, is secretly a group of permutations.
Every group is secretly a permutation group. Even $\mathbb{Z}$, which lives in $S_\mathbb{Z}$, which is uncountable. Cool. Cool cool cool.
If $G$ is a group then it is isomorphic to a subgroup of $S_G$.
The idea: each element $g$ acts on the group by left multiplication, shuffling the elements around. That shuffle is a permutation of $G$, and multiplying elements corresponds to composing their shuffles.
Proof. $\forall g \in G$ define $\tau_g : G \rightarrow G$ by $\tau_g(h) = gh$.
Claim: $\tau_g \in S_G$.
Proof of claim:
Therefore, $\tau_g$ is a bijection from $G$ to $G$.
Now define $\phi : G \rightarrow S_G$ by $\phi(g) = \tau_g$. Then:
It follows that $\phi(G) \le S_G$ (Property 2), and that the map $\tilde{\phi} : G \rightarrow \phi(G)$ defined by $\tilde{\phi}(g) = \phi(g)$ is an isomorphism. Therefore $G \cong \phi(G)$. $\blacksquare$
Notice that this $\tau_g$ is left multiplication, not the conjugation map from Example 8. Below, each group is drawn as a ring of its elements, and $\tau_g$ is drawn as arrows $h \mapsto gh$. The third ring checks the key line of the proof: doing $\tau_{g_2}$ and then $\tau_{g_1}$ is the same as doing $\tau_{g_1 g_2}$.
Pick a group, then pick $g_1$ and $g_2$. Each ring shows a left multiplication map as a permutation of the elements of $G$.
Play with $S_3$ for a bit. When $g_1 g_2 \neq g_2 g_1$, the permutations $\tau_{g_1}$ and $\tau_{g_2}$ don't commute either. That's exactly what an injective homomorphism should do: it copies the whole multiplication table, non-commuting pairs included.
Let $V_4 = \{e, a, b, c\}$ be the Klein 4-group. Label its elements $e = 1$, $a = 2$, $b = 3$, $c = 4$, and write each $\tau_g$ ($g \in V_4$) from Cayley's theorem as an element of $S_4$ in cycle notation.
In $V_4$, every element is its own inverse and the product of two different non-identity elements is the third. So:
So $V_4 \cong \{e, (1 \ 2)(3 \ 4), (1 \ 3)(2 \ 4), (1 \ 4)(2 \ 3)\} \le S_4$.
How do you actually find homomorphisms? Two facts do most of the work: a homomorphism can't increase the order of an element, and it's pinned down by what it does to generators. First, orders.
If $g \in G$ and $|g| < \infty$ then $|\phi(g)| \mid |g|$.
The last step is the lemma from Cyclic Groups: if $|h| = m$, then $h^k = e$ exactly when $m \mid k$.
Proof. Suppose $|g| = n \in \mathbb{N}$. Then $e_H = \phi(e_G) = \phi(g^n) = \phi(g)^n$ (by 1a and 1c) $\Rightarrow |\phi(g)| \mid n$. $\blacksquare$
Now generators. If you know where the generators go, you know where everything goes.
If $G = \langle S \rangle$ then the homomorphism $\phi$ is completely determined by $\{\phi(s) : s \in S\}$. In other words, if $\tilde{\phi} : G \rightarrow H$ is any homomorphism with $\tilde{\phi}(s) = \phi(s)$, $\forall s \in S$, then $\tilde{\phi} = \phi$.
Proof. Suppose $\tilde{\phi} : G \rightarrow H$ is a homomorphism with $\tilde{\phi}(s) = \phi(s)$, $\forall s \in S$. Then $\forall g \in G$, $\exists n \in \mathbb{N}$, $g_1, \dots, g_n \in S$, and $u_1, \dots, u_n \in \{\pm 1\}$ s.t. $g = g_1^{u_1} g_2^{u_2} \cdots g_n^{u_n}$. Then
$$\begin{align*} \tilde{\phi}(g) &= \tilde{\phi}(g_1)^{u_1} \tilde{\phi}(g_2)^{u_2} \cdots \tilde{\phi}(g_n)^{u_n} \\ &= \phi(g_1)^{u_1} \phi(g_2)^{u_2} \cdots \phi(g_n)^{u_n} = \phi(g). \quad \blacksquare \end{align*}$$
That's a uniqueness statement. It does not say every choice of values works.
Important: When defining a homomorphism $\phi : G \rightarrow H$, just choosing values for $\phi(s)$, for $s \in S$, doesn't guarantee that $\phi$ will actually extend to a homomorphism on all of $G$. The examples below show choices that look perfectly reasonable and still fail.
Find all homomorphisms $\phi : \mathbb{Z}/8\mathbb{Z} \rightarrow C_6$.
Write $C_6 = \langle x \rangle = \{e, x, x^2, x^3, x^4, x^5\}$. Since $\mathbb{Z}/8\mathbb{Z} = \langle 1 \rangle$, any homomorphism $\phi : \mathbb{Z}/8\mathbb{Z} \rightarrow C_6$ is determined by $\phi(1)$. There are six choices to consider:
So, there are 2 homomorphisms from $\mathbb{Z}/8\mathbb{Z}$ to $C_6$: the trivial homomorphism, and the homomorphism determined by the rule $1 \mapsto x^3$.
The "another way to see this" argument is worth watching in motion. Walk $k = 0, 1, 2, \dots$ around $\mathbb{Z}/m\mathbb{Z}$ and $\phi(k) = x^{jk}$ around $C_n$ at the same time. When $k$ wraps back around to $0$, $\phi(k)$ had better be back at $e$.
Choose $m$, $n$, and where the generator goes, then press WALK. In $S_3$ mode, choose where both generators go and check each relation.
With two generators, there's more to go wrong.
Find all homomorphisms $\phi : S_3 \rightarrow C_6$.
Write $S_3 = \langle (1 \ 2), (1 \ 2 \ 3) \rangle$ (orders $2$ and $3$), and $C_6 = \langle x \rangle = \{e, x, x^2, x^3, x^4, x^5\}$ (orders $1, 6, 3, 2, 3, 6$).
Any homomorphism is determined by $\phi((1 \ 2))$ and $\phi((1 \ 2 \ 3))$. Since $|\phi(g)| \mid |g|$, we must have $\phi((1 \ 2)) = e$ or $x^3$, and $\phi((1 \ 2 \ 3)) = e$, $x^2$, or $x^4$ (6 possibilities).
So, there are 2 homomorphisms from $S_3$ to $C_6$: the trivial homomorphism, and the homomorphism determined by the rule $(1 \ 2) \mapsto x^3$, $(1 \ 2 \ 3) \mapsto e$.
The failures in Example 13 all had the same flavour: some equation that's true in $S_3$ (like $(2 \ 3)^2 = e$) became false after applying $\phi$. The next property says that's the only thing that can go wrong.
Suppose that $G = \langle S \rangle$, that $\phi : S \rightarrow H$, and that any relation satisfied by elements $s_1, \dots, s_n \in S$ is also satisfied with these replaced by $\phi(s_1), \dots, \phi(s_n)$. Then $\phi$ extends uniquely to a homomorphism from $G$ to $H$.
Checking "any relation" sounds impossible, since there are infinitely many. Luckily, a presentation does the heavy lifting.
When a presentation for $G$ is given (i.e. using generators and relations), you only need to check the above condition for the relations in the presentation.
That turns proving two groups isomorphic into a short checklist.
On Symmetric Groups we proved $S_3 \cong D_6$ by elimination: every group of order $6$ is $C_6$ or $D_6$, and $S_3$ isn't Abelian. Here's a direct proof that builds the isomorphism.
Show that $D_6 \cong S_3$.
$D_6 = \langle r, s \mid r^3 = s^2 = e, rs = sr^{-1} \rangle$. Let $\phi(r) = (1 \ 2 \ 3)$, $\phi(s) = (1 \ 2)$. Then:
Therefore, $\phi$ extends to a homomorphism $\phi : D_6 \rightarrow S_3$. But $3 = |\phi(r)| \mid |\phi(D_6)|$ and $2 = |\phi(s)| \mid |\phi(D_6)|$ (Lagrange's theorem) $\Rightarrow |\phi(D_6)| = 6 \Rightarrow \phi$ is bijective, therefore it is an isomorphism.
Homomorphisms that are isomorphisms are the start of a much bigger story: how to tell when two groups are the same, and how to prove they aren't.
Isomorphisms preserve everything about a group's structure: orders of elements, being Abelian, the whole lattice of subgroups. On Isomorphism Theorems, homomorphisms and quotient groups turn out to be two views of the same thing.
Find all homomorphisms $\phi : \mathbb{Z}/6\mathbb{Z} \rightarrow \mathbb{Z}/4\mathbb{Z}$.
$\mathbb{Z}/6\mathbb{Z} = \langle 1 \rangle$, so $\phi$ is determined by $\phi(1)$. By Property 5, $|\phi(1)|$ divides $|1| = 6$. It also divides $4$ (Lagrange's theorem in $\mathbb{Z}/4\mathbb{Z}$), so $|\phi(1)| \in \{1, 2\}$, which leaves $\phi(1) = 0$ or $\phi(1) = 2$.
So there are exactly 2 homomorphisms. (Try $m = 6$, $n = 4$ in the Wrap-Around Test.)
Example 12 found $2$ homomorphisms $\mathbb{Z}/8\mathbb{Z} \rightarrow C_6$, and the last exercise found $2$ from $\mathbb{Z}/6\mathbb{Z}$ to $\mathbb{Z}/4\mathbb{Z}$. Notice that $(8, 6) = (6, 4) = 2$. Coincidence?
Let $m, n \in \mathbb{N}$. Prove that there are exactly $(m, n)$ homomorphisms from $\mathbb{Z}/m\mathbb{Z}$ to $\mathbb{Z}/n\mathbb{Z}$.
Proof. Since $\mathbb{Z}/m\mathbb{Z} = \langle 1 \rangle$, a homomorphism is determined by $a = \phi(1) \in \{0, 1, \dots, n-1\}$ (Property 6a), and then $\phi(k) = ka \pmod n$ (by 1c, written additively). So we need to count the values of $a$ that actually give a homomorphism.
Necessary: if $\phi$ is a homomorphism then $0 = \phi(0) = \phi(m \cdot 1) = ma \pmod n$, so $n \mid ma$.
Sufficient: if $n \mid ma$, define $\phi(k) = ka \pmod n$. This is well defined: if $k \equiv k' \pmod m$ then $k - k' = tm$ for some $t \in \mathbb{Z}$, so $ka - k'a = t(ma) \equiv 0 \pmod n$. And $\phi(k + l) = (k + l)a = ka + la = \phi(k) + \phi(l)$, so $\phi$ is a homomorphism.
Counting: write $d = (m, n)$. Then $n \mid ma \iff \frac{n}{d} \mid \frac{m}{d} a$. Since $\left(\frac{n}{d}, \frac{m}{d}\right) = 1$, this happens if and only if $\frac{n}{d} \mid a$. The multiples of $\frac{n}{d}$ in $\{0, 1, \dots, n - 1\}$ are $0, \frac{n}{d}, 2\frac{n}{d}, \dots, (d-1)\frac{n}{d}$, and there are exactly $d$ of them. $\blacksquare$
In particular, if $(m, n) = 1$, the only homomorphism from $\mathbb{Z}/m\mathbb{Z}$ to $\mathbb{Z}/n\mathbb{Z}$ is the trivial one.