= Solution
Multiplicativity of the <determinant> gives $\delta(gh)=\delta(g)\delta(h)$ and $\delta(1)=1$, so $\delta$ is a <linear character>. Its finite image is a cyclic <group> of <roots of unity>. If it contains $-1$, its order is $2m$ and the <character of a representation> $\delta^m$ maps onto $\{\pm1\}$. The <kernel of a group homomorphism> is the preimage of its identity, so $\ker(\delta^m)$ is a <normal subgroup> of index $2$. The <subgroup> $\ker\delta$ itself need not have index $2$.
By <Cauchy theorem for groups>, a <group> $E$ of order $2k$ has an involution $g$. In the <regular representation>, left multiplication by $g$ is a permutation consisting of $k$ disjoint transpositions, so its <determinant> is $(-1)^k=-1$ when $k$ is odd. The preceding argument gives \b[a <normal subgroup> of index two].
For a nonabelian <simple group> of order below $80$, <Burnside's theorem> requires at least three distinct <prime> factors. An odd order with three factors is at least $3\cdot5\cdot7=105$. An even order with exactly one factor of $2$ has a <normal subgroup> of index $2$. Thus its order must be divisible by $4$ and by at least two odd <primes>. Below $80$ the only possibility is $4\cdot3\cdot5=60$, since the next is $4\cdot3\cdot7=84$. Hence $\boxed{|H|=60}$.
Back to article page
