UPSC Maths 2025 Paper 2 Q1b — Solution

10 marks · Section A

Question

Let $G = \{e, x, x^2, y, yx, yx^2\}$ be a non-Abelian group with $o(x) = 3$ and $o(y) = 2$ . Show that $xy = yx^2$ (where $e$ is the identity element of $G$ and $o(x)$ , $o(y)$ denote the order of the elements $x$ , $y$ respectively).

Technique

Examine the conjugate $yxy^{-1}$ in this order-6 group (the dihedral group $D_3 \cong S_3$ ); conjugation preserves order, which pins $yxy^{-1}$ down to one of two values, and the non-Abelian hypothesis eliminates the wrong one.

Solution

We have a group $G$ of order 6 generated by $x$ (order 3) and $y$ (order 2), with the six distinct elements listed. Since $o(y)=2$ , we have $y^{-1}=y$ .

Step 1 — Consider the conjugate $yxy^{-1} = yxy$ .

Since the six listed elements are all of $G$ , the element $yxy$ must equal one of $\{e, x, x^2, y, yx, yx^2\}$ . Conjugation preserves order, so $o(yxy^{-1}) = o(x) = 3$ . Thus $yxy$ is an element of order 3. The only elements of $G$ of order 3 are $x$ and $x^2$ (here $o(e)=1$ , $o(y)=2$ , and $y, yx, yx^2$ are the three reflections, each of order 2 in $D_3$ ). Hence $yxy \in \{x, x^2\}.$

Step 2 — Rule out $yxy = x$ .

If $yxy = x$ , then $yx = xy$ , so $x$ and $y$ commute. But then every product of powers of $x$ and $y$ commutes, making $G$ Abelian — contradicting the hypothesis. Therefore $yxy \neq x$ , so $yxy = x^2.$

Step 3 — Derive $xy = yx^2$ .

From $yxy = x^2$ , left-multiply both sides by $y$ (using $y^2=e$ ): $y(yxy) = yx^2 \implies (y^2)xy = yx^2 \implies xy = yx^2.$

This is exactly the required relation. $\qquad\blacksquare$

Consistency check. With $x^3=e$ , $y^2=e$ , $yx=x^2y$ , the six elements $e,\ x,\ x^2,\ y,\ yx,\ yx^2$ are closed under multiplication and form the dihedral group $D_3\cong S_3$ , with the relation $xy=yx^2$ governing all such reductions.

Answer

$\boxed{\,xy = yx^2\,}$ established from $yxy^{-1}=x^2$ (forced because conjugation preserves the order 3, and $yxy^{-1}=x$ would make $G$ Abelian).