Isomorphism theorems (First, Second, Third)

At a Glance

• Frequency: 2 sub-parts across 2 of 13 years (2018, 2022)
• Priority tier: T3
• Marks (count): 15 (2)
• Average solve time: ~8 min
• Difficulty mix: easy 2
• Section: A | Dominant type: proof

Why This Chapter Matters

Both UPSC questions on the First Isomorphism Theorem are 15 marks, making this one of the highest-value single-theorem items in Paper 2 Algebra. The 2018 question applies the theorem to identify $\mathbb R/\mathbb Z\cong S^1$; the 2022 question asks for the complete proof. Both reduce to the same four-step structure: define a surjective homomorphism, compute its kernel, invoke FIT, state the conclusion. Knowing the proof cold and being able to exhibit a concrete surjection on demand covers both archetypes with one preparation.

Minimum Theory

First Isomorphism Theorem. Let $\phi:G\to G'$ be a group homomorphism. Let $K=\ker\phi$. Then: (i) $K\trianglelefteq G$; (ii) there is an isomorphism $\bar\phi:G/K\xrightarrow{\;\sim\;}\operatorname{Im}\phi$ defined by $\bar\phi(gK)=\phi(g)$. In particular, $G/\ker\phi\cong\operatorname{Im}\phi$.

Well-definedness check (the crucial step). $\bar\phi$ is well-defined because: if $gK=g'K$ then $g^{-1}g'\in K$, so $\phi(g^{-1}g')=e$, so $\phi(g)=\phi(g')$.

Second and Third Isomorphism Theorems (stated for reference). Second: If $H\le G$ and $N\trianglelefteq G$, then $H\cap N\trianglelefteq H$ and $H/(H\cap N)\cong HN/N$. Third: If $N\trianglelefteq G$ and $K\trianglelefteq G$ with $N\le K$, then $K/N\trianglelefteq G/N$ and $(G/N)/(K/N)\cong G/K$.

Unit circle group. $S^1=\{z\in\mathbb C:|z|=1\}$ under multiplication. The map $x\mapsto e^{2\pi ix}$ sends $(\mathbb R,+)$ onto $(S^1,\times)$ with kernel $\mathbb Z$.

Question Archetypes

first-isomorphism-application (1 question(s); 2018)

Recognition Cues — “Show $G/N\cong H$” where $H$ is a known group; “quotient group is isomorphic to …”; find a natural surjective homomorphism $\phi:G\to H$ with $\ker\phi=N$.

Solution Template

1. Define $\phi:G\to H$ and verify it is a homomorphism.
2. Show $\phi$ is surjective.
3. Compute $\ker\phi$ and verify $\ker\phi = N$ (the named subgroup).
4. Apply FIT: $G/N = G/\ker\phi\cong\operatorname{Im}\phi = H$.

Worked Example

2018 Paper 2, 2018-P2-Q2a (15 marks)

Show that the quotient group $(\mathbb R,+)/\mathbb Z$ is isomorphic to the multiplicative group of complex numbers on the unit circle.

Let $S^1=\{z\in\mathbb C:|z|=1\}$. Define $\phi:\mathbb R\to S^1$ by $\phi(x)=e^{2\pi ix}$.

Step 1 — Homomorphism. $\phi(x+y)=e^{2\pi i(x+y)}=e^{2\pi ix}\cdot e^{2\pi iy}=\phi(x)\phi(y)$. $\checkmark$

Step 2 — Surjectivity. For any $e^{i\theta}\in S^1$, set $x=\theta/(2\pi)\in\mathbb R$; then $\phi(x)=e^{i\theta}$. $\checkmark$

Step 3 — Kernel. $\phi(x)=1\iff e^{2\pi ix}=1\iff 2\pi x = 2\pi n$ for some $n\in\mathbb Z\iff x\in\mathbb Z$. So $\ker\phi=\mathbb Z$.

Step 4 — First Isomorphism Theorem. $\phi$ is a surjective homomorphism with kernel $\mathbb Z$, so

$\mathbb R/\mathbb Z = \mathbb R/\ker\phi \;\cong\; \operatorname{Im}\phi = S^1.$

$\boxed{\mathbb R/\mathbb Z\cong S^1.}\;\blacksquare$

(Well-definedness on cosets: if $x'=x+n$ ($n\in\mathbb Z$), then $e^{2\pi ix'}=e^{2\pi ix}e^{2\pi in}=e^{2\pi ix}$, so the induced map $\bar\phi(x+\mathbb Z)=e^{2\pi ix}$ is independent of the coset representative.)

Common Traps

• Use $e^{2\pi ix}$ (period 1) so the kernel is exactly $\mathbb Z$. Using $e^{ix}$ gives kernel $2\pi\mathbb Z$ and quotient $\mathbb R/2\pi\mathbb Z\cong S^1$ instead — same conclusion, but the kernel does not match $\mathbb Z$.
• Verifying surjectivity (every point of $S^1$ is hit) and computing the kernel are both required for full marks.
• State the First Isomorphism Theorem explicitly by name; don’t just say “therefore the map descends.”

first-isomorphism-proof (1 question(s); 2022)

Recognition Cues — “Prove every homomorphic image of $G$ is isomorphic to some quotient group of $G$”; “State and prove the First Isomorphism Theorem.”

Solution Template

1. State the theorem: $\phi:G\to G'$ a homomorphism, $K=\ker\phi$; then $K\trianglelefteq G$ and $G/K\cong\operatorname{Im}\phi$.
2. Prove $K\le G$: identity, closure, inverses.
3. Prove $K\trianglelefteq G$: for any $g\in G$, $k\in K$: $\phi(gkg^{-1})=\phi(g)\cdot e\cdot\phi(g)^{-1}=e$, so $gKg^{-1}\subseteq K$.
4. Define $\bar\phi:G/K\to\operatorname{Im}\phi$ by $\bar\phi(gK)=\phi(g)$.
5. Check well-definedness, homomorphism, surjectivity, injectivity.
6. Conclude $\bar\phi$ is an isomorphism.

Worked Example

2022 Paper 2, 2022-P2-Q2b (15 marks)

Prove that every homomorphic image of a group $G$ is isomorphic to some quotient group of $G$.

Let $\phi:G\to G'$ be a homomorphism. Set $K=\ker\phi=\{g\in G:\phi(g)=e_{G'}\}$.

Step 1 — $K$ is a subgroup. $e\in K$ (since $\phi(e)=e'$). If $a,b\in K$: $\phi(ab)=\phi(a)\phi(b)=e'\cdot e'=e'$, so $ab\in K$. If $a\in K$: $\phi(a^{-1})=\phi(a)^{-1}=(e')^{-1}=e'$, so $a^{-1}\in K$. $\checkmark$

Step 2 — $K\trianglelefteq G$. For $g\in G$, $k\in K$: $\phi(gkg^{-1})=\phi(g)\phi(k)\phi(g)^{-1}=\phi(g)\cdot e'\cdot\phi(g)^{-1}=e'$. So $gkg^{-1}\in K$, i.e. $gKg^{-1}\subseteq K$. $\checkmark$

Step 3 — Define $\bar\phi:G/K\to\operatorname{Im}\phi$ by $\bar\phi(gK)=\phi(g)$.

Well-defined: if $gK=g'K$, then $g^{-1}g'\in K$, so $\phi(g^{-1}g')=e'$, so $\phi(g)=\phi(g')$. $\checkmark$

Homomorphism: $\bar\phi(gK\cdot g'K)=\bar\phi(gg'K)=\phi(gg')=\phi(g)\phi(g')=\bar\phi(gK)\bar\phi(g'K)$. $\checkmark$

Surjective: for any $y\in\operatorname{Im}\phi$, $y=\phi(g)=\bar\phi(gK)$. $\checkmark$

Injective: if $\bar\phi(gK)=e'$, then $\phi(g)=e'$, so $g\in K$, so $gK=K$ (identity in $G/K$). Hence $\ker\bar\phi=\{K\}$. $\checkmark$

Conclusion. $\bar\phi$ is a bijective homomorphism, i.e. an isomorphism.

$\boxed{G/K\cong\operatorname{Im}\phi,\quad K=\ker\phi.}\;\blacksquare$

Common Traps

• Well-definedness is the most commonly incomplete step. Write it out: ”$gK=g'K\Rightarrow g^{-1}g'\in K\Rightarrow\phi(g^{-1}g')=e'\Rightarrow\phi(g)=\phi(g')$.”
• Normality of $K$ must be proved, not assumed; injectivity of $\bar\phi$ follows from $\ker\bar\phi=\{K\}$.
• Both surjectivity (onto $\operatorname{Im}\phi$, not onto $G'$) and injectivity must be checked for $\bar\phi$ to be an isomorphism.

Marks-Aware Writing

Both questions are 15 marks. For the application question (2018): the four steps (define $\phi$, verify homomorphism, verify surjectivity, compute kernel, cite FIT) together cover all marks — missing the surjectivity check or the kernel computation each costs 3 marks. For the proof question (2022): Steps 2 (normality) and 3 (well-definedness + four properties of $\bar\phi$) are the load-bearing sections. A complete proof of normality plus a complete four-part check of $\bar\phi$ earns 12–15 marks; a proof that omits well-definedness or injectivity earns at most 9.

Practice Set

• 2024-P2-Q2b (15 m) — — Hint: prove that commutativity transfers to the image via the homomorphism property; then exhibit $S_3\to\mathbb Z_2$ as a counterexample for the converse.