In article <3753386.1208278400069.JavaMail.jakarta@[EMAIL PROTECTED]
>,
allpro <nowimpsbball@[EMAIL PROTECTED]
> wrote:
>Let |G| = (p^n)m where p is prime and gcd(p,m) = 1. Suppose that H is
>a normal subgroup of G of order p^n. If K is a subgroup of G of order
>p^k, show that K is subgroup of H.
>
>Okay, I wonder if there is more I need to do, or if I need to prove
>they are finite.
If you need to prove that ->who<- is finite? G is given to be finite,
H is given to be finite, and K is given to be finite, so who are you
having doubts about?
>I feel like I am missing something...but here is
>what I got
>p^k has to be less than p^n because if p^k was bigger than p^n then
>p^k would not divide the order of G because p and m are relatively
>prime and K could not be a subgroup of G. The order of a subgroup
>must divide the order of the group.
Yes.
>Both H and K are subgroups of G, they both are closed under the same
>operation as G, and because n>k, p^k divides p^n and thus because K
>is closed under the operation of H
This does not follow: you cannot say that K is "closed under the
operation of H" unless you can show first that the restriction of the
operation of H to K makes sense, and that will only make sense if you
first show that K is a subgroup of H... which is what you are supposed
to prove in the first place.
But this is completely unnecessary anyway: the order of K must divide
the order of G by Lagrange's Theorem. So p^k must divide p^n*m. Since
m is relatively prime to p, it follows that p^k must divide p^n, and
therefore that k is less than or equal to n (you should not exclude the
case that k=n).
>and K's order divides the order of
>H, K must be a subgroup of H.
No, this is false as well. It is NOT true in general that if G is a
group, K and H are subgroups of G, H is normal, and the order of K
divides the order of H, then K is a subgroup of H. For example, take G
to be the Klein 4-group, G = C_2 x C_2, take H to be C_2 x {0}, and
take K to be {0} x C_2. (Of course, it does not satisfy the full
properties of what you want to prove, but the point is that the
general argument you are claiming is in fact false in general).
Do you know Sylow's Theorems yet? What you want to prove follows
easily from them.
--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
======================================================================
Arturo Magidin
magidin-at-member-ams-org
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