Past exam of the short options course of the University of Oxford / 2014 / 2 / c

Solution

588 = 2^{2} \cdot 3 \cdot 7^{2}

. The Sylow theorems say that the number of Sylow subgroups for the prime seven divides

12

and is congruent to one modulo seven. Among

1, 2, 3, 4, 6, 12

, only one qualifies. Thus there is a unique normal subgroup

N

of order

49

Every group of prime-square order is Abelian. Indeed its center of a group is nontrivial, and a cyclic quotient by the center of a group forces the whole group to be abelian. Consequently

N

is either

C_{49}

C_{7} \times C_{7}

, and in either case it has two composition factors, both

C_{7}

It remains to consider

Q = H / N

, of order twelve. Its number of Sylow subgroups for the prime three is either one or four:

If it is one, the normal order-three subgroup has factor $C_{3}$ , and the remaining quotient has order four. An order-four group is abelian and has two factors $C_{2}$ .
If it is four, these order-three subgroups have disjoint nonidentity elements and account for eight elements. Any Sylow subgroup for the prime two, of order four, must consist of the four remaining elements, since its nonidentity elements have power-of-two orders. It is therefore unique and normal. Its two factors are $C_{2}$ , and the quotient has order three, giving $C_{3}$ .

Q

H

N

. This gives

C_{2}, C_{2}, C_{3}, C_{7}, C_{7}

as the multiset of composition factors. The Jordan–Hölder theorem makes this independent of the chosen series. In particular every group of order

588

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