Past exam of the mathematics course of the University of Cambridge / 2024 / ib / Paper 1 / 9E (Groups, Rings and Modules)

Let $|G|=p^{a}m$ with $p\nmid m$. The Sylow theorems state:

Since $|S_3|=6$, its Sylow $2$-subgroups have order two and its Sylow $3$-subgroups have order three. They areandrespectively. The latter is the unique Sylow $3$-subgroup.

A Sylow $2$-subgroup of $S_4$ has order $8$. For example,It is the normalizer of $⟨(1234)⟩$. The six $4$-cycles form three inverse pairs, and a dihedral group of order eight has a unique cyclic subgroup of order four. Hence the three Sylow $2$-subgroups are

A Sylow $2$-subgroup of $A_5$ has order four. Since $A_5$ has no element of order four, each is a Klein four-group. For every $i\in \{1,\dots ,5\}$, take the identity and the three double transpositions fixing $i$. For instance, the subgroup fixing $5$ isThese five point stabilizers are precisely the Sylow $2$-subgroups. Indeed, the $15$ double transpositions occur three to each such subgroup, so there areof them. Together with the previous parts, this gives the Sylow subgroups of S3, S4 and A5.

Let $G$ act by left multiplication on the coset set $G/H$. Composing the resulting homomorphismwith the sign homomorphism gives a homomorphism $G\to \{\pm 1\}$. If it were nontrivial, its kernel would have index two, contrary to the hypothesis. Thus every element of $G$, and in particular $x$, induces an even permutation.

Since $P$ is a Sylow $2$-subgroup, $[G:P]$ is odd. ThereforeAn involution acts as disjoint transpositions and fixed points. If $x$ fixed no coset, it would be a product of $[G:H]/2$ transpositions, an odd number, contradicting evenness. HenceThis is the even-involution coset fixed-point lemma.

If $gH$ is a fixed coset, thenMultiplying on the left by $g^{-1}$ givesso $g^{-1}xg\in H$. Thus