Homomorphisms and Isomorphisms

Let $\varphi: \mathbb Z \to \mathbb Z_5$ send $k$ to its remainder mod $5$. Check it on $a = 8, b = 19$: $\varphi(8) = 3$ and $\varphi(19) = 4$. Directly, $\varphi(8 + 19) = \varphi(27) = 2$, and going the other way, $\varphi(8) + \varphi(19) = 3 + 4 = 7 \equiv 2 \pmod 5$ as well — the two routes to the answer agree, as the homomorphism equation demands for every pair, not just this one.

For $\varphi: GL_2(\mathbb Q) \to \mathbb Q^{\times}$ given by $\varphi(M) = \det M$, take $A = \begin{pmatrix}1&2\\0&1\end{pmatrix}$ and $B = \begin{pmatrix}1&0\\3&1\end{pmatrix}$. Then $AB = \begin{pmatrix}7&2\\3&1\end{pmatrix}$, so $\det(AB) = 7 - 6 = 1$, while $\det A \cdot \det B = 1 \cdot 1 = 1$: they agree. $\det(AB) = \det A \det B$ is a theorem of linear algebra, and it says exactly that $\det$ carries matrix multiplication to real multiplication — the homomorphism property in disguise.

A homomorphism $\varphi: \mathbb Z_6 \to \mathbb Z_4$ is determined by $\varphi(1)$, since every element of $\mathbb Z_6$ is a multiple of the generator $1$. Applying $\varphi$ to $6 \cdot 1 = 0$ forces $6\varphi(1) \equiv 0 \pmod 4$, so $\varphi(1)$ must have order dividing $\gcd(6, 4) = 2$ in $\mathbb Z_4$. Only $0$ and $2$ have order dividing $2$, giving exactly $2$ homomorphisms: the trivial map $\varphi(k) = 0$, and $\varphi(k) = 2k \bmod 4$. Checking the second one stays consistent all the way around: $\varphi(1) = 2, \varphi(2) = 0, \varphi(3) = 2, \varphi(4) = 0, \varphi(5) = 2, \varphi(6) = \varphi(0) = 0$.

Let $\varphi: \mathbb Z_8 \to \mathbb Z_4$ send $k$ to $k \bmod 4$. Then $\ker\varphi = \{0, 4\}$, since those are exactly the elements sent to $0$, so $|\ker\varphi| = 2$. Every residue $0, 1, 2, 3$ is hit (by $k$ and $k + 4$ alike), so $\operatorname{im}\varphi = \mathbb Z_4$ and $|\operatorname{im}\varphi| = 4$. Notice $|\ker\varphi| \cdot |\operatorname{im}\varphi| = 2 \cdot 4 = 8 = |\mathbb Z_8|$ — no coincidence; it previews the counting identity behind the First Isomorphism Theorem.

Let $\varphi: \mathbb Z \to \mathbb Z$ send $k$ to $3k$; it is a homomorphism of $(\mathbb Z, +)$ since $3(a + b) = 3a + 3b$. If $3k = 0$ then $k = 0$, so $\ker\varphi = \{0\}$ and $\varphi$ is injective. But $\operatorname{im}\varphi = 3\mathbb Z = \{\ldots, -3, 0, 3, 6, \ldots\}$, not all of $\mathbb Z$: for instance $1 \notin \operatorname{im}\varphi$, since $3k = 1$ has no integer solution. Injective does not imply surjective.

For the sign homomorphism $\operatorname{sgn}: S_3 \to \{1, -1\}$, the kernel is $\ker(\operatorname{sgn}) = \{e, (1\,2\,3), (1\,3\,2)\}$ — the even permutations, also called $A_3$ — since those are the only elements of $S_3$ with sign $+1$. The image is all of $\{1, -1\}$, because the transposition $(1\,2)$ has sign $-1$. So $|\ker(\operatorname{sgn})| = 3$ and $|\operatorname{im}(\operatorname{sgn})| = 2$, and again $3 \cdot 2 = 6 = |S_3|$.

Define $\varphi: \mathbb Z_6 \to \mathbb Z_2 \times \mathbb Z_3$ by $\varphi(k) = (k \bmod 2, k \bmod 3)$. Listing all six images: $\varphi(0) = (0,0), \varphi(1) = (1,1), \varphi(2) = (0,2), \varphi(3) = (1,0), \varphi(4) = (0,1), \varphi(5) = (1,2)$ — six distinct pairs, exactly the six elements of $\mathbb Z_2 \times \mathbb Z_3$, so $\varphi$ is a bijection. It is a homomorphism because reduction mod $2$ and mod $3$ are each homomorphisms; for instance $\varphi(2 + 5) = \varphi(1) = (1,1)$ matches $\varphi(2) + \varphi(5) = (0,2) + (1,2) = (1, 4 \bmod 3) = (1,1)$.

Both have order $4$, but check element orders in $U(8) = \{1,3,5,7\}$ under multiplication mod $8$: $3^2 = 9 \equiv 1, 5^2 = 25 \equiv 1, 7^2 = 49 \equiv 1 \pmod 8$, so every non-identity element has order $2$ — $U(8)$ has no element of order $4$. But $\mathbb Z_4$ has one (namely $1$). Since isomorphisms preserve element order, $U(8) \not\cong \mathbb Z_4$: same size, different internal structure.

In $U(5) = \{1,2,3,4\}$, the powers of $2$ are $2^1 = 2, 2^2 = 4, 2^3 = 3, 2^4 = 1 \pmod 5$, so $2$ generates all of $U(5)$: it is cyclic of order $4$. The map $\varphi(2^k \bmod 5) = k \bmod 4$ turns multiplication into addition of exponents, so it is an isomorphism $U(5) \cong \mathbb Z_4$. Check one instance: $2 \cdot 2 = 4 = 2^2$ corresponds to $\varphi(2) + \varphi(2) = 1 + 1 = 2$.

Take $r = (1\,2\,3)$ and $s = (1\,2)$ in $S_3$. Composing right-to-left, $(rs)(1) = r(s(1)) = r(2) = 3$, while $(sr)(1) = s(r(1)) = s(2) = 1$. Since $3 \ne 1$, $rs \ne sr$: $S_3$ is non-abelian. But $\mathbb Z_6$ is abelian — it is cyclic — and "abelian" is an isomorphism invariant, so $S_3 \not\cong \mathbb Z_6$, even though both groups have order $6$.

Build $D_4$, the symmetries of a square, as $\{e, r, r^2, r^3, s, rs, r^2s, r^3s\}$ with $r^4 = s^2 = e$ and $sr = r^{-1}s$. Computing orders: $r$ and $r^3$ have order $4$, while $r^2, s, rs, r^2s, r^3s$ all have order $2$ — five elements of order $2$ in total. In $\mathbb Z_8$, only the element $4$ has order $2$ (since $\operatorname{ord}(k) = 8/\gcd(k,8)$). Both groups have order $8$, but they disagree on the number of elements of order $2$ ($5$ versus $1$), so $\mathbb Z_8 \not\cong D_4$ — confirmed without ever listing a full bijection.

There are exactly two groups of order $4$ up to isomorphism: $\mathbb Z_4$, which has an element of order $4$, and $\mathbb Z_2 \times \mathbb Z_2$ (the Klein four-group $V_4$), in which every non-identity element has order $2$ since $2(a,b) = (0,0)$ for all $a, b \in \mathbb Z_2$. The presence or absence of an order-$4$ element separates them instantly, so $\mathbb Z_4 \not\cong V_4$.

Every automorphism of $\mathbb Z_6$ has the form $\varphi_a(k) = ak \bmod 6$ for a unit $a \in U(6) = \{1, 5\}$, so $|\operatorname{Aut}(\mathbb Z_6)| = 2$. The nontrivial one is $\varphi_5(k) = 5k \bmod 6$. Check it is a homomorphism on $2, 3$: $\varphi_5(2) + \varphi_5(3) = 4 + 3 = 7 \equiv 1 \pmod 6$, matching $\varphi_5(2 + 3) = \varphi_5(5) = 25 \bmod 6 = 1$.

As the $U(8)$ table above showed, the units mod $8$ are $\{1, 3, 5, 7\}$, each of order at most $2$: in particular $3^2 = 9 \equiv 1 \pmod 8$, so the automorphism $\varphi_3$ satisfies $\varphi_3 \circ \varphi_3 = \varphi_{9 \bmod 8} = \varphi_1 = \operatorname{id}$. So $\operatorname{Aut}(\mathbb Z_8) \cong U(8)$ has order $4$ but every non-identity element squares to the identity — it is the Klein four-group from the Invariants section, not $\mathbb Z_4$.

Conjugation by a fixed element is always an automorphism: $\varphi_x(g) = xgx^{-1}$. In $S_3$, conjugating the $3$-cycle $(1\,2\,3)$ by the transposition $(1\,2)$ relabels its entries by swapping $1 \leftrightarrow 2$: $(1\,2\,3)$ becomes $(2\,1\,3) = (1\,3\,2)$, the inverse $3$-cycle. So $\varphi_{(1\,2)}$ is a nontrivial inner automorphism of $S_3$, even though $S_3$ has only $6$ elements.