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Let G be a group of order 24 that is not isomorphic to S4. Then one of it

Let G be a group of order 24 that is not isomorphic to

S_4

. Then one of its Sylow subgroups is normal.

This is the proof from my textbook (http://www.albany.edu/~mark/algebra.pdf).

Proof

Suppose that the 3-Sylow subgroups are not normal. The number of 3-Sylow
subgroups is 1 mod 3 and divides 8. Thus, if there is more than one 3-Sylow subgroup,
there must be four of them.

Let X be the set of 3-Sylow subgroups of G. Then G acts on X by conjugation, so we
get a homomorphism

f : G \rightarrow S(X) \cong S_4

. As we’ve seen in the discussion on G-sets, the kernel of f is the intersection of the isotropy subgroups of the elements of X. Moreover, since the action is that given by conjugation, the isotropy subgroup of

H \in X

N_G(H)

(the normalizer of H in G). Thus,

ker f = \cap_{H \in X} N_G(H).

For

H \in X

, the index of

N_G(H)

is 4, the number of conjugates of H. Thus, the order of

N_G(H)

is 6. Suppose that K is a different element of X. We claim that the order of

N_G(H) \cap N_G(K)

divides 2.

To see this, note that the order of

N_G(H) \cap N_G(K)

cannot be divisible by 3. This is because any p-group contained in the normalizer of a p-Sylow subgroup must be contained in the p-Sylow subgroup itself (Corollary 5.3.5). Since the 3-Sylow subgroups have prime order here, they cannot intersect unless they are equal. But if the order of

N_G(H) \cap N_G(K)

divides 6 and is not divisible by 3, it must divide 2.

In consequence, we see that the order of the kernel of f divides 2. If the kernel has
order 1, then f is an isomorphism, since G and

S_4

have the same number of elements.

Thus, we shall assume that ker f has order 2. In this case, the image of f has order
12. But by Problem 2 of Exercises 4.2.18,

A_4

is the only subgroup of

S_4

of order 12, so we must have im f =

A_4

.

By Problem 1 of Exercises 4.2.18, the 2-Sylow subgroup,

P_2

, of

A_4

is normal. But since ker f has order 2,

f^{−1}P_2

has order 8, and must be a 2-Sylow subgroup of G. As the pre-image of a normal subgroup, it must be normal, and we’re done.

My Question

I'm just confused about the last part. I kind of got lost when it was explaining how/why

Unparseable latex formula:

f^{-1}P_2$

has order 8. I'm not really sure how that's related to the kernel of f.

Thank you in advance

(edited 10 years ago)

Reply 1

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