Normal Subgroup with Prime Index (3.3.3)

Dummit and Foote section 3.3, exercise 3:

Let G be a group, let H be a normal subgroup of G with | G : H | = p for some prime p. Then, for all K≤G:

(i) K≤H, or
(ii) HK=G and | K : H∩K | = p

Proof: If K is not a subgroup of H, then that implies that there exists k∈K, k∉H. This then implies that kH≠H within G/H. Since G/H forms a group itself, this implies that < kH > is a subgroup of G/H. But G/H is a group of prime order, p, so any subgroup of G/H must divide p, so | kH | = 1 or p. Since the former implies that kH=H, | kH | = p.

But G/H is only composed of p elements/cosets, and since these cosets partition G, that implies that G is formed in entirety by the union of the elements of < kH >; therefore, KH=HK=G.

Now, since K ≤ N_KH(H), the Second Isomorphism theorem implies:

(KH)/H ≅ K/(K∩H) ⇒
G/H ≅ K/(H∩K) ⇒
| G : H | = | K : H∩K | ⇒
| K : H∩K | = p

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