IE 33
(Another example of this phenomenon is that all automorphisms of the
symmetric group Sn are inner for all integers n 1 except n 6.)
1.32. Theorem. Let \G\
p3q,
where p and q are primes. Then G has
either a normal Sylow p-subgroup or a normal Sylow q-subgroup, except when
\G\ = 24.
Proof. As in the previous proof, we assume that np 1 and that nq 1,
and we conclude that p ^ q and that np p and nq q. Continuing to
reason as we did previously, we have nq q = np p. Since nq divides
p3,
we are left this time with two possibilities: either nq =
p3
or nq =
p2.
Suppose that nq =
p3.
Then the
p3
subgroups of order q contain a total of
p3(q—
1) elements of order #, and thus G contains at most |G|
p3(q
1) =
p3
elements that do not have order q. It follows that if P G Sylp(G), then P is
exactly the set of elements of G having order different from g, and thus P
is unique, contradicting the assumption that np 1.
We conclude, therefore, that nq =
p2,
and thus
p2
= 1 mod q. In other
words, q divides
p2
1 = (p + l)(p—l). Since q is prime, we see that q must
divide either p+1 or p 1, and so in either case, we have q p + 1. But
we know from the first paragraph of the proof that p q, and so the only
possibility is that q = p + 1. Now p and q are primes that differ by 1, and
so one of them must be even. We deduce that p = 2 and q = 3, and thus
|G| - 24.
Of course, the previous proof does not tell us that there actually is an
exception of order 24; it only permits the possibility of such an exception.
But an exception really does exist. Consider 54, the symmetric group on
four symbols, which, of course, has order 4! = 24. By examining the possible
cycle structures for elements of 54, it is easy to count that there are exactly
nine elements of order 2, eight elements of order 3 and six elements of order
4. (As a check, note that 9+8+6 = 23, so that together with the identity, we
have accounted for all 24 elements.) By the Sylow D-theorem, every element
of 2-power order in 54 lies in some Sylow 2-subgroup. Since each Sylow 2-
subgroup has order 8, and yet there are more than eight elements of 2-power
order, it follows that 54 must have more than one Sylow 2-subgroup. Similar
reasoning shows that there must be more that one Sylow 3-subgroup. (In
fact, it is easy to see that n2(54) 3 and that 723(54) = 4.)
There is still more that we can say about this situation. We know
that among groups of order
p3q,
only groups of order 24 can fail to have a
nontrivial normal Sylow subgroup, and we know that at least one group of
order 24 actually does fail to have such a Sylow subgroup. The remaining
question is whether or not there are any other groups of order 24 that can
occur as exceptions. The answer is "no".
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