36 1. Sylow Theory
the n!-theorem by taking P G Syl2(G), so that \G : P\ = 3. This completes
the proof that 60 is the smallest possible order of a nonabelian simple group.
To prove that no number between 60 and 168 can be the order of a
simple group, it is convenient to have two more nonsimplicity results of the
type we have been discussing: if G has order
or order pqr, where p, q
and r are primes, then G cannot be simple. (We leave the proofs of these
to the exercises at the end of this section.) Using these two facts and the
other results that we have established, all but about a dozen of the numbers
between 60 and 168 can be eliminated, and most of those are easily dealt
with by ad hoc methods. For example, we need to consider the numbers
84 =
140 =
and 156 =
and these are easily eliminated
by the observation that if p is the largest prime divisor, then np 1 since
no divisor of n exceeding 1 is congruent to 1 modulo p.
The number 72 =
can be eliminated by observing that if G is
simple and |G| = 72, then n% 1, and therefore the only possibility is
ri3 = 4. Then \G : N\ = 4, where N is the normalizer of a Sylow 3-
subgroup, and this violates the n!-theorem. More interesting is the case
where \G\ = 144 =
This case can be handled using Theorem 1.16,
with an argument analogous to the one we used in the discussion following
Corollary 1.17, where we showed that groups of order
are not simple.
Another number that requires some work is 132 =
If a group of this
order is simple, then nu 12, and hence there are at least 120 elements of
order 11, leaving at most 12 other elements. It follows that ns = 4, and this
yields a contradiction using the n!-theorem.
The number between 60 and 168 that seems to be the most difficult to
eliminate is 120 =
One (rather tricky) approach is this. If G is simple
of order 120, it is easy to see that n^(G) = 6. Then G acts (nontrivially)
on the six right cosets of the normalizer of a Sylow 5-subgroup, and this
determines a nontrivial homomorphism from G into the symmetric group
SQ. Since G is simple, this homomorphism is injective, and thus there is an
isomorphic copy H of G with H C SQ. But G has no normal subgroup of
index 2, and hence by Lemma 1.34, no element of G can act oddly on the
six cosets, and thus H actually lies in the alternating group AQ, and since
\AQ\ 360, we see that 1^6 : H\ 3. Recall, however, that the alternating
groups An are simple when n 5, and so by the n!-theorem, AQ cannot
have a subgroup of index 3. This contradiction shows that 120 is not the
order of a simple group.
An alternative approach for eliminating 120 is to quote Problem 1C.4,
which asserts that if \G\ = 120, then there exists a subgroup H C G with
l | G : i J | 5 . IfGi s simple, this yields an isomorphism of G into the
symmetric group 55. Since IS5I = 120 = |G|, we conclude that G = S$,
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