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Let be the action of on the 2-Sylow subgroups. Then , order 24, acts on of order 6, implying a normal subgroup in of order 4. This subgroup then combines with a 3-Sylow subgroup, by semi-direct product, to form a subgroup of of order 12. There can be only one such subgroup of order 12, as there is no room for any more.
2. Since normal Sylow subgroups are unique, if X and Y are 3 -Sylows and X ≠ Y, then N(X) ≠ N(Y) and N(X) ∩ N(Y) is a 3′ -group, and so has order 1 or 2 as each normalizer has order 6. If their intersection has order 1 then the action of G on its four 3 -Sylow subgroups is faithful and so since |G| = 24, G must be isomorphic to S4.
2. Z24 Z 24 is the only cyclic one - all other groups are non-abelian except Z12 ×Z2 Z 12 × Z 2, which you noted is not cyclic. Among the remaining ones, Z12 ×Z2 Z 12 × Z 2 is the only abelian one, so this is out too. If you have already seen the center of a group, you may proceed as follows. Among the remaining ones, S4 S 4 is the only ...
Group of order. 24. is not simple. How can we show that a group of order 24 is not simple? This is my attempt: | G | = 24 = 3 ⋅ 23 so nG(2) ∈ {1, 3} where nG(p) donotes the number of Sylow p -groups. If nG(2) = 1 we are done. Otherwise consider the action (by conjugation) of G on Syl 2(G) the Sylow 2-groups. This action is transitive since ...
6. A simple group G G of order 24 24 would have three Sylow 2 2 -subgroups, and so there would be a non-trivial homomorphism from G G to S3 S 3 (see the comment) which will be injective by simplicity and therefore 24 24 would divide 6 6 which is absurd. Because if it was not injective, it would have a non-trivial kernel, which would be a non ...
I have slightly expanded my comment above into an answer. Let G G be a group of order 24 24, let T T be a Sylow 3 3 -subgroup, and let n3 n 3 be the number of Sylow 3 3 -subgroups. Sylow's theorems imply that n3 n 3 is 1 1 or 4 4. If n3 = 4 n 3 = 4, then the normaliser of T T has index 4 4. If n3 = 1 n 3 = 1, then T T is normal in G G, and G/T ...
1 Answer. Sorted by: 3. By Lagrange's theorem, order of a a is a divisor of 24 24. Since a12 a 12 and a8 a 8 are not identity, this order does not divide 12 12 or 8 8. Consider the divisors of 24 24 and eliminate each divisor that divides 12 12 or 8 8. You are left with a single divisor which must be the order of a a.
So the groups can have either 1 or 4 sylow 3 subgroups of order 3, and either 1 or 3 sylow 2 subgroups of order 4. Furthermore coprime sylow-p groups only share the identity element in common, so we can rule out 4 sylow-3 groups and 3 sylow-2 groups, as the presence of either rules out the existence of a single copy of the other.
Prove a group of order $22,000$ with $16$ Sylow-$5$ subgroup has a normal (Sylow) $11$ subgroup. Hot Network Questions Water is coming up from my vinyl flooring, near the toilet and across the bathroom floor in my new home.
Another example is the dihedral group of order $24$, which can be obtained as a semidirect product of the cyclic groups of order $2$ and order $12$. (Recall that cyclic groups are abelian.) (Recall that cyclic groups are abelian.)