Prove that A5 has a subgroup of order 12 Solution suppose, by way of contradicti

Question

Prove that A5 has a subgroup of order 12

Explanation / Answer

suppose, by way of contradiction, that G had such a subgroup, H. let G/H be the set of left cosets of H in G. by assumption, there are less than 5 such cosets. define f:G -->Sym(G/H) by f(g):xH-->gxH. a)f is a homomorphism: f(gg') is the element that takes xH-->gg'xH f(g)o°(g') takes xH--->g'xH-->g(g'xH) = gg'xH. thus f(gg') = f(g)of(g'), f is a homomorphism. since f is a homomorphism, ker(f) is a normal subgroup of G. since G is simple, we have 2 choices: ker(f) = {e}, or ker(f) = G. if ker(f) = G, then every g yields the identity map on G/H: xH --> xH, for all xH in G/H. so, in particular, we must have gH = H, for all g in G, which is not possible, since H is not all of G. therefore, ker(f) = {e}. so f(G) is isomorphic to G, so |f(G)| = |G|. but this means |G| divides ([G:H])!, which is at most 4! = 24. so, we are reduced to showing that no non-abelian group of orders 1,2,3,4,6,8,12 and 24 are simple. any group of order 1,2 or 3 is cyclic, hence abelian. any group of order 4 is abelian. the only non-abelian group of order 6 is S3, which has the normal subgroup A3. if |G| = 24, G is isomorphic to S4 by our argument above, and S4 has the normal subgroup A4. that leaves |G| = 8, and |G| = 12. if |G| = 12, then G is (isomorphic to) a subgroup of order 12 of S4. the only such subgroup is A4, which has the normal subgroup V = {e, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3) }. that leaves |G| = 8. there are 2 non-abelian groups of order 8, D4 and Q8. D4 has the normal subgroup {1,r,r^2,r^3}, while Q8 has the normal subgroup {1,-1,i,-i} (alternately, any group of order 8 is a p-group, with p = 2, so has a non-trivial center. since G is assumed non-abelian, Z(G) is a non-trivial normal subgroup, so G cannot be simple). since there is no simple subgroup G with order dividing 24, we conclude that any non-abelian simple group, cannot have a subgroup with index < 5. (NOTE: technically this only shows this is impossible if [G:H] = 4. but if [G:H] = 3, we have by the same argument G is isomorphic to a subgroup of S3, and S3 has no non-abelian subgroups. if [G:H] = 2, then H is normal, contradicting the fact that G is simple).

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