If G is a finite Abelian group, H, K ≤ G, and if x ∈ H K with o(x) = p a prime, show that H or K must contain an element of order p.

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- March 6th 2013, 11:21 AMhenderson7878abstract algebra: subgroups question
If G is a finite Abelian group, H, K ≤ G, and if x ∈ H K with o(x) = p a prime, show that H or K must contain an element of order p.

- March 6th 2013, 11:35 AMjakncokeRe: abstract algebra: subgroups question
can you tell me what is? is it an element of inner automorphisms of (HK) ?

- March 6th 2013, 01:52 PMILikeSerenaRe: abstract algebra: subgroups question
Hi henderson7878! :)

There are 3 cases:

1. x in H

2. x in K

3. x neither in H nor in K, and x = hk

What conclusions can you draw in each of these cases for x^{p}?

(o(x) is a notation for the order of x.) - March 6th 2013, 02:01 PMhenderson7878Re: abstract algebra: subgroups question
o(x) is the order

- March 8th 2013, 03:37 PMjohngRe: abstract algebra: subgroups question
Here is a stronger statement:

Let G be a finite group, H and K subgroups of G. Assume the product set HK = { hk : h is in H and k is in K } is also a subgroup of G. (If G is abelian, this is true for any two subgroups H, K.) Let p be a prime and hk in HK of order p. Then either H contains an element of order p or K contains an element of order p.

Proof. Since hk is in HK, p divides the order of HK. . (Actually, this is true even without the supposition that HK is a subgroup. I leave the proof to you.) So p divides |H| or p divides |K|. By Cauchy's Theorem, one of H and K contain an element of order p.

Extra Credit: Show by counterexample, this statement is false if HK is not a subgroup of G. Hint - consider the alternating group A_{4}. - March 9th 2013, 01:46 PMNehushtanRe: Abstract algebra: Subgroups question
Let , , and suppose .

Then (as is Abelian) . Thus divides and so either or .

Suppose divides , say . Then . Moreover as . Hence has order .

Similarly if divides then has order .