Past exam of the mathematics course of the University of Cambridge / 2011 / ia / Paper 3 / 5D (Groups)

The order of a group element $g$ is the least positive integer $m$ for which $g^m=e$, if one exists. In a finite group, two terms of $e,g,g^2,\dots ,g^{|G|}$ must agree. If $g^i=g^j$ with $i<j$, cancellation gives $g^{j-i}=e$. Thus such positive integers exist and have a least member: every element of a finite group has finite order.

If $h=ag{a}^{-1}$, repeated cancellation shows $h^m=a{g}^m{a}^{-1}$ for every positive integer $m$. Consequently $h^m=e$ exactly when $g^m=e$. The least positive exponents are therefore equal:This proof also shows that conjugation preserves infinite order when the ambient group is not finite.

For a permutation $g$, start with a letter $i$ and follow $i,g(i),g^2(i),\dots$. Finiteness gives a repetition; since $g$ is invertible, the first return is to $i$. The visited letters form a permutation cycle on which $g$ acts by moving to the next letter. If there are letters left, repeat the construction starting with one of them. Its trajectory cannot meet an earlier one, because applying a suitable inverse power of $g$ would put its starting letter in that earlier trajectory. Continuing partitions the letters into disjoint cycles, including fixed points as cycles of length one.

The resulting disjoint permutation cycles have disjoint supports, and their product agrees with $g$ on every letter. This proves the cycle decomposition of a permutation. A power $g^m$ fixes a cycle of length $l$ exactly when $l\mid m$, so if the cycle lengths are $l_1,\dots ,l_t$,The least common multiple is essential: multiplying the lengths would overcount whenever cycles share a factor.

The alternating group is the subgroup of the symmetric group consisting of even permutations:For a permutation cycle of length $l$,using rightmost-first composition. Its sign of a permutation is therefore $(-1{)}^{l-1}$. Multiplicativity of the sign homomorphism givesThusEquivalently, $n-t$ is even when $t$ counts all cycles, including fixed points. Odd-length cycles contribute positive sign and impose no further restriction.

Extend $\sigma \in S_n$ to a permutation $\hat{\sigma }$ of $n+2$ letters that fixes the last two letters. Let $t=(n+1n+2)$, and let $ϵ(\sigma )$ be $0$ or $1$ according as $\sigma$ is even or odd. DefineThe two factors have disjoint supports and commute. Their signs cancel, so $\iota (\sigma )\in A_{n+2}$. Additivity of parity of a permutation modulo two givesThus $\iota$ is a group homomorphism. Its restriction to the first $n$ letters recovers $\sigma$, making it injective. Its image is therefore a subgroup isomorphic to $S_n$:This group embedding works for every positive $n$, including the trivial case $n=1$.