Proofs : Grorups, Subgroups and Indexes
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1.) If |G|=p^n for some prime p, and 1 is not equal to H, a normal
subgroup of G, show that H intersect Z(G) is not equal to 1.
2.) Suppose G is finite, H is a subgroup of G, [G:H]=n and |G| does not
divide n!. Show that there is a normal subgroup K of G, with K not
equal to 1, such that K is a subgroup of H.
3.) Prove that if G contains no subgroup of index 2, then any subgroup
of index 3 is normal in G.
4.) Suppose G acts on S, x in G, and x in S. Show that
Stab_G(xs)=xStab_G(s)x^-1.
5.) Let G be a group of order 15, which acts on a set S with 7 elements
Show the group action has a fixed point.
6.)a.) Suppose that H and K both have finite index in G. Prove that
[G:H intersect K]is a subgroup of [G:H][G:K].
b.) Suppose that [G:H] is finite. Prove that H contains a normal
subgroup of G which, in turn, has finite index in G.
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(1) Recall that if G is a p-group acting on a finite set X, then
|X_G| = |X| mod p
where X_G = {x in X : gx = x for all g in G}.
Now suppose that H is a nontrivial normal subgroup of a p-group G; then G acts on H by conjugation, so
H_G = {h in H : xgx^{-1} = g for all g in G} = H n Z(G).
From the above remark, it follows |H n Z(G)| = |H| mod p; and since H is nontrivial, neither is H n Z(G), as desired.
(2) As before, G acts on the left cosets of H, and this action induces a homomorphism G --> S_n, with kernel K.
If K is trivial, then |G| divides n!, contrary to assumption, so K cannot be trivial. ...
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