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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- Mar 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.

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

- Mar 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.) - Mar 6th 2013, 02:01 PMhenderson7878Re: abstract algebra: subgroups question
o(x) is the order

- Mar 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. $\displaystyle |HK|=|H|\times{|K|\over |H\cap K|}$. (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}. - Mar 9th 2013, 01:46 PMNehushtanRe: Abstract algebra: Subgroups question
Let $\displaystyle x=hk$, $\displaystyle h\in H,k\in K$, and suppose $\displaystyle \mathrm o(h)=a,\mathrm o(k)=b$.

Then $\displaystyle x^{ab}$ $\displaystyle =$ $\displaystyle (hk)^{ab}$ $\displaystyle =$ $\displaystyle h^{ab}k^{ab}$ (as $\displaystyle G$ is Abelian) $\displaystyle =$ $\displaystyle e$. Thus $\displaystyle p=\mathrm o(x)$ divides $\displaystyle ab$ and so either $\displaystyle p\mid a$ or $\displaystyle p\mid b$.

Suppose $\displaystyle p$ divides $\displaystyle a$, say $\displaystyle a=cp$. Then $\displaystyle e=h^a=(h^c)^p$. Moreover $\displaystyle h^c\ne e$ as $\displaystyle c<a=\mathrm o(h)$. Hence $\displaystyle h^c\in H$ has order $\displaystyle p$.

Similarly if $\displaystyle p$ divides $\displaystyle b$ then $\displaystyle k^{b/p}\in K$ has order $\displaystyle p$.