Group homomorphisms: kernel, image

At a Glance

• Frequency: 4 sub-parts across 4 of 13 years (2019, 2020, 2023, 2024)
• Priority tier: T3
• Marks (count): 10 (3), 15 (1)
• Average solve time: ~6 min
• Difficulty mix: easy 4
• Section: A | Dominant type: proof

Why This Chapter Matters

Group homomorphism questions are the most frequent algebra topic, appearing in four separate years for a combined 45 marks. All three 10-mark variants follow the same two-line proof structure: show the image order divides both $|G|$ and $|H|$, then use coprimality or the normal subgroup classification to rule out non-trivial images. The 15-mark question asks you to prove that properties like commutativity transfer to homomorphic images. Mastering the core lemma (image order divides domain order via the First Isomorphism Theorem) unlocks all four archetypes at once.

Minimum Theory

Group homomorphism. A map $\phi:G\to H$ is a group homomorphism if $\phi(ab)=\phi(a)\phi(b)$ for all $a,b\in G$. The kernel is $\ker\phi=\{g\in G:\phi(g)=e_H\}$ and the image is $\operatorname{Im}\phi=\phi(G)=\{\phi(g):g\in G\}$.

Basic properties. (i) $\ker\phi\trianglelefteq G$ (normal subgroup); (ii) $\operatorname{Im}\phi\le H$ (subgroup); (iii) $\phi$ is injective iff $\ker\phi=\{e_G\}$.

First Isomorphism Theorem. $G/\ker\phi\cong\operatorname{Im}\phi$. In particular, $|\operatorname{Im}\phi|=|G|/|\ker\phi|$, so $|\operatorname{Im}\phi|$ divides $|G|$. Combined with Lagrange in $H$: $|\operatorname{Im}\phi|$ divides $|H|$ as well. Hence $|\operatorname{Im}\phi|$ divides $\gcd(|G|,|H|)$.

Question Archetypes

homomorphism-existence (3 question(s); 2019, 2020, 2023)

Recognition Cues — “prove the only homomorphism is trivial”; “show there is no homomorphism of $G$ onto $H$”; given concrete groups ($S_3$, $\mathbb Z_3$, groups of given orders).

Solution Template

1. Let $\phi:G\to H$ be any homomorphism.
2. Use FIT: $|\operatorname{Im}\phi|$ divides $|G|$ (via $G/\ker\phi$) and divides $|H|$ (Lagrange in $H$).
3. Hence $|\operatorname{Im}\phi|$ divides $\gcd(|G|,|H|)$.
4. If $\gcd(|G|,|H|)=1$: conclude $|\operatorname{Im}\phi|=1$, so $\phi$ is trivial.
5. If asking about surjectivity: $|\operatorname{Im}\phi|=|H|$ requires $|H|\mid|G|$; check whether this holds.
6. For specific groups (e.g. $S_3\to\mathbb Z_3$): use the abelianization argument — any map to an abelian group kills commutators, constraining the image order further.

Worked Example

2019 Paper 2, 2019-P2-Q2a (10 marks)

If $G$ and $H$ are finite groups whose orders are relatively prime, prove that there is only one homomorphism from $G$ to $H$, the trivial one.

Let $\phi:G\to H$. By the First Isomorphism Theorem, $\operatorname{Im}\phi\cong G/\ker\phi$, so $|\operatorname{Im}\phi|$ divides $|G|$. By Lagrange in $H$, $|\operatorname{Im}\phi|$ divides $|H|$. Hence

$|\operatorname{Im}\phi| \;\Big|\; \gcd(|G|,|H|) = 1.$

So $|\operatorname{Im}\phi|=1$, meaning $\phi(g)=e_H$ for all $g\in G$. Since $\phi$ was arbitrary, the trivial map is the only homomorphism $G\to H$.

$\boxed{\gcd(|G|,|H|)=1\;\Longrightarrow\;\text{the only homomorphism }G\to H\text{ is }g\mapsto e_H.}\;\blacksquare$

(Element-order version: for any $g\in G$, $\operatorname{ord}(\phi(g))$ divides both $\operatorname{ord}(g)\mid|G|$ and $|H|$. Coprimality forces $\operatorname{ord}(\phi(g))=1$, so $\phi(g)=e_H$.)

2020 Paper 2, 2020-P2-Q1a (10 marks)

Show that there is no homomorphism of $S_3$ into $\mathbb Z_3$ except the trivial homomorphism.

Let $\phi:S_3\to\mathbb Z_3$. Since $\mathbb Z_3$ is abelian, $\phi$ factors through the abelianization $S_3/[S_3,S_3]=S_3/A_3\cong\mathbb Z_2$. So $|\operatorname{Im}\phi|$ divides $|S_3/A_3|=2$.

But also $\operatorname{Im}\phi\le\mathbb Z_3$, so $|\operatorname{Im}\phi|$ divides $3$.

Since $\gcd(2,3)=1$, $|\operatorname{Im}\phi|=1$, i.e. $\phi$ is trivial. $\blacksquare$

(Direct route via FIT: $|\operatorname{Im}\phi|$ divides $|S_3|=6$ and $|\mathbb Z_3|=3$, so $|\operatorname{Im}\phi|\in\{1,3\}$. If $|\operatorname{Im}\phi|=3$, then $|\ker\phi|=2$ — a normal subgroup of $S_3$ of order 2. But $S_3$ has no normal subgroup of order 2 (its only normal subgroups are $\{e\}$, $A_3$, $S_3$). Contradiction. So $|\operatorname{Im}\phi|=1$.)

$\boxed{\text{Every homomorphism }S_3\to\mathbb Z_3\text{ is trivial.}}\;\blacksquare$

2023 Paper 2, 2023-P2-Q1a (10 marks)

Let $G$ be a group of order 10 and $G'$ be a group of order 6. Examine whether there exists a homomorphism of $G$ onto $G'$.

Suppose $\phi:G\twoheadrightarrow G'$ is surjective. By FIT, $G/\ker\phi\cong G'$, so $|G'|=|G|/|\ker\phi|$ divides $|G|=10$. But $6\nmid 10$ (since $10=2\cdot5$ lacks the factor $3$). Contradiction.

$\boxed{\text{No surjective homomorphism from }G\text{ to }G'\text{ exists.}}\;\blacksquare$

Common Traps

• The image order divides $|G|$ (via FIT) as well as $|H|$ — forgetting the $|G|$ side and only using $|\operatorname{Im}\phi|\mid|H|$ misses the coprimality argument.
• For surjectivity: the constraint is $|G'|\mid|G|$ (not $|G|\mid|G'|$). Confusing the direction loses the contradiction.
• The trivial homomorphism $g\mapsto e_H$ always exists; the question often asks whether a non-trivial or onto one exists.
• For the $S_3\to\mathbb Z_3$ case: ruling out $|\operatorname{Im}\phi|=3$ requires explicitly invoking the classification of normal subgroups of $S_3$.

homomorphic-image-property (1 question(s); 2024)

Recognition Cues — “Show every homomorphic image of a [property] group is [property]”; “converse is not necessarily true”; needs a counterexample.

Solution Template

1. Let $\phi:G\to H$ be a homomorphism and let $G$ have property $P$.
2. Take arbitrary elements $h_1,h_2\in\operatorname{Im}\phi$; lift them to $g_1,g_2\in G$ with $\phi(g_i)=h_i$.
3. Use property $P$ on $g_1,g_2$ in $G$, then apply $\phi$ to transfer the property to $h_1,h_2$.
4. For the converse: exhibit a non-$P$ group $G$ with a homomorphism $\phi:G\to H$ where $\operatorname{Im}\phi$ has property $P$.

Worked Example

2024 Paper 2, 2024-P2-Q2b (15 marks)

Show that every homomorphic image of an abelian group is abelian, but the converse is not necessarily true.

Part 1. Let $\phi:G\to H$ be a homomorphism and $G$ abelian. Take $h_1,h_2\in\operatorname{Im}\phi$; pick $g_1,g_2\in G$ with $\phi(g_i)=h_i$. Then

$h_1h_2=\phi(g_1)\phi(g_2)=\phi(g_1g_2)=\phi(g_2g_1)=\phi(g_2)\phi(g_1)=h_2h_1.$

So $\operatorname{Im}\phi$ is abelian. $\square$

Part 2 — Converse fails. Take $G=S_3$ (non-abelian: $(12)(13)=(132)\ne(123)=(13)(12)$) and the sign homomorphism $\phi:S_3\to\mathbb Z_2$, $\phi(\sigma)=\operatorname{sgn}(\sigma)$. This is a homomorphism with image $\mathbb Z_2=\{+1,-1\}$, which is abelian. But $G=S_3$ is not abelian.

$\boxed{\text{Homomorphic image of abelian is abelian; converse fails (witness: }S_3\to\mathbb Z_2).}\;\blacksquare$

Common Traps

• “Homomorphic image” is $\operatorname{Im}\phi=\phi(G)$, not the entire codomain $H$. The proof uses elements of $\operatorname{Im}\phi$ only.
• The converse counterexample requires a non-abelian $G$ whose image under some $\phi$ is abelian — $S_3/A_3\cong\mathbb Z_2$ is the cleanest choice.
• In Part 1, the step $\phi(g_1g_2)=\phi(g_2g_1)$ uses commutativity in $G$; make this explicit.

Marks-Aware Writing

For 10-mark existence questions: a complete answer has (a) FIT cited by name, (b) the two-sided divisibility argument for $|\operatorname{Im}\phi|$, (c) the numerical conclusion ruling out non-trivial images, and (d) a boxed statement. Stating FIT without computing $|\ker\phi|$ earns only 3–4 marks.

For the 15-mark property-transfer question: Part 1 needs the lifted-element argument written out symbol by symbol (4–5 marks), and Part 2 needs a specific, verified counterexample — naming $S_3\to\mathbb Z_2$ without checking $S_3$ is non-abelian loses 2 marks.

Practice Set

• 2022-P2-Q2b (15 m) — — Hint: prove the First Isomorphism Theorem directly; check well-definedness of the induced map.
• 2022-P2-Q1a (10 m) — — Hint: apply FIT; compute divisibility constraints on the image order.