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UPSC 2020 Maths Optional Paper 2 Q1a — Step-by-Step Solution

10 marks · Section A

Group homomorphisms: kernel, image · Algebra · asked 4× in 13 yrs · Read the full method →

Question

Let $S_3$ and $Z_3$ be the permutation group on 3 symbols and the group of residue classes modulo 3 respectively. Show that there is no homomorphism of $S_3$ into $Z_3$ except the trivial homomorphism.

Technique

First Isomorphism Theorem + Lagrange + classification of normal subgroups of $S_3$; alternative via abelianization $S_3/[S_3,S_3]\cong Z_2$ and coprimality $\gcd(2,3)=1$.

Solution

Let $\varphi:S_3\to Z_3$ be any group homomorphism (here $Z_3=(\mathbb{Z}/3\mathbb{Z},+)$ is the additive cyclic group of order $3$). We show $\varphi$ is trivial, i.e. $\varphi(g)=0$ for all $g\in S_3$.

Step 1 — Use the First Isomorphism Theorem

Let $K=\ker\varphi\trianglelefteq S_3$. By the First Isomorphism Theorem,

Hence $|\operatorname{im}\varphi|$ divides $|Z_3|=3$ (Lagrange), so $|\operatorname{im}\varphi|\in\{1,3\}$.

Also, by Lagrange applied to $S_3$,

so $|\operatorname{im}\varphi|$ divides $6$. Combined with dividing $3$, this allows $|\operatorname{im}\varphi|\in\{1,3\}$ a priori — we must rule out $3$.

Step 2 — The normal subgroups of $S_3$

If $|\operatorname{im}\varphi|=3$ then $|K|=6/3=2$, so $K$ would be a normal subgroup of order $2$.

The subgroups of order $2$ in $S_3$ are exactly the three generated by a single transposition: $\langle(1\,2)\rangle,\ \langle(1\,3)\rangle,\ \langle(2\,3)\rangle$. None of these is normal: e.g.

Thus $S_3$ has no normal subgroup of order $2$. (Indeed the only normal subgroups of $S_3$ are $\{e\}$, the alternating group $A_3$ of order $3$, and $S_3$ itself.)

Therefore $|K|=2$ is impossible, so $|\operatorname{im}\varphi|\ne 3$.

Step 3 — Conclude

The only remaining possibility is $|\operatorname{im}\varphi|=1$, i.e. $\operatorname{im}\varphi=\{0\}$ and $K=S_3$. Hence $\varphi(g)=0$ for every $g\in S_3$: $\varphi$ is the trivial homomorphism.

Answer