← 2022 Paper 2

UPSC 2022 Maths Optional Paper 2 Q2b — Step-by-Step Solution

15 marks · Section A

Isomorphism theorems (First, Second, Third) · Algebra · asked 2× in 14 yrs · Read the full method →

Question

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

Technique

Standard proof of First Isomorphism Theorem: kernel is normal; define $\bar\phi$ on cosets; show well-defined, homomorphism, bijective.

Solution

Statement (First Isomorphism Theorem). Let $\phi:G\to G'$ be a group homomorphism with kernel $K=\ker\phi=\{g\in G:\phi(g)=e_{G'}\}$. Then:

1. $K$ is a normal subgroup of $G$.
2. $G/K\cong\phi(G)$ (the image).

So the homomorphic image $\phi(G)$ is isomorphic to the quotient $G/K$.

Step 1 — $K$ is a subgroup

• $e\in K$ since $\phi(e)=e_{G'}$.
• 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 $a^{-1}\in K$.

$K$ is a subgroup ✓.

Step 2 — $K$ is normal

For any $g\in G$ and $k\in K$: $\phi(gkg^{-1})=\phi(g)\phi(k)\phi(g^{-1})=\phi(g)\cdot e\cdot\phi(g)^{-1}=e$.

So $gkg^{-1}\in K$. Hence $gKg^{-1}\subseteq K$ for all $g\in G$ ⇒ $K\triangleleft G$.

Step 3 — Define isomorphism $\bar\phi:G/K\to\phi(G)$

Define $\bar\phi(gK)=\phi(g)$.

Well-defined. If $gK=g'K$, then $g^{-1}g'\in K$, so $\phi(g^{-1}g')=e$, i.e., $\phi(g)^{-1}\phi(g')=e$, i.e., $\phi(g)=\phi(g')$. So $\bar\phi$ is well-defined.

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

Surjective. For any $y\in\phi(G)$, $y=\phi(g)$ for some $g\in G$, so $y=\bar\phi(gK)$.

Injective. If $\bar\phi(gK)=e_{G'}$, then $\phi(g)=e$, so $g\in K$, so $gK=K$ = identity in $G/K$. So $\ker\bar\phi=\{K\}$, hence $\bar\phi$ is injective.

Step 4 — Conclusion

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

Answer