Question

Prove that if a group g of order 28 has a normal subgroup of order 4 then g is abelian

Answer 1

Answer:

Let G be a group such that | G|=28. ... that |H|=4, and H is a normal subgroup of G. Previously, I have ... the order of elements in G? When H is C4, G will be C 28 ...

Answer 2

Let G be a group such that |G|=28. We are given H, such that |H|=4, and H is a normal subgroup of G. Previously, I have proven that G must also contain a normal subgroup, K, where |K|=7. This was done without Sylow. H must be isomorphic to C4 or C2×C2 because these are the only groups of order 4, up to isomorphism. G is abelian, I guessed that G will be C28 or C2×C14.

Using the Direct Product Theorem. In either case of the identity of H, H∩K=e, because H will not contain any elements of order 7, and all the elements of K are order 7 apart from the identity. Also, H and K are normal, so their elements commute with each other: For h∈H and k∈K, (khk−1)h−1∈H and k(hk−1h−1)∈K means khk−1h−1=e.

When H is C4, G will be C28 and so must contain an element of order 28.

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