# abstract algebra: subgroups question

• Mar 6th 2013, 11:21 AM
henderson7878
abstract 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 AM
jakncoke
Re: abstract algebra: subgroups question
can you tell me what $o(x)$ is? is it an element of inner automorphisms of (HK) ?
• Mar 6th 2013, 01:52 PM
ILikeSerena
Re: 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 xp?

(o(x) is a notation for the order of x.)
• Mar 6th 2013, 02:01 PM
henderson7878
Re: abstract algebra: subgroups question
o(x) is the order
• Mar 8th 2013, 03:37 PM
johng
Re: 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. $|HK|=|H|\times{|K|\over |H\cap K|}$. (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 A4.
• Mar 9th 2013, 01:46 PM
Nehushtan
Re: Abstract algebra: Subgroups question
Let $x=hk$, $h\in H,k\in K$, and suppose $\mathrm o(h)=a,\mathrm o(k)=b$.

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

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

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