= Solution
Compose <permutations> from right to left. Tracking the images gives $1\mapsto2\mapsto1$, $3\mapsto4\mapsto3$, and $5\mapsto5$, so
$$
\boxed{(123)(234)=(12)(34).}
$$
Two <transpositions> give <sign of a permutation> $+1$, proving that this element belongs to the <alternating group> $A_5$.
Its <conjugacy class> consists of all double transpositions. Conjugation merely relabels the letters, so no other cycle type can occur. Conversely, choose a permutation $h\in S_5$ relabelling the two pairs into any desired two pairs. If $h$ is odd, replace it by $h(12)$: the transposition $(12)$ centralizes $(12)(34)$, so this replacement is even and gives the same conjugate. Thus every double transposition is reached by conjugation inside $A_5$.
The complete list of fifteen elements, grouped by their fixed letter, is
$$
\begin{aligned}
&(12)(34),\ (13)(24),\ (14)(23),\\
&(12)(35),\ (13)(25),\ (15)(23),\\
&(12)(45),\ (14)(25),\ (15)(24),\\
&(13)(45),\ (14)(35),\ (15)(34),\\
&(23)(45),\ (24)(35),\ (25)(34).
\end{aligned}
$$
There are five choices of fixed letter and three pairings of the other four letters, confirming the <conjugacy class> size $15$.
Back to article page
