On Tue, 15 Apr 2008 12:52:48 EDT, allpro
<nowimpsbball@[EMAIL PROTECTED]
> wrote in
<news:3753386.1208278400069.JavaMail.jakarta@[EMAIL PROTECTED]
>
in alt.math.undergrad:
> 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. I feel like I am missing
> something...but here is what I got
Prove that *what* are finite? You know that G, H, and K are
finite, because you know their orders.
> 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.
Actually, all you know is that p^k <= p^n, so that k <= n;
you don't know that p^k is strictly smaller than p^n.
> Both H and K are subgroups of G, they both are closed
> under the same operation as G, and because n>k,
No, n >= k. But it's still true that p^k | p^n.
> p^k divides p^n and thus because K is closed under the
> operation of H and K's order divides the order of H, K
> must be a subgroup of H.
This makes no sense at all, I'm afraid. It is not true in
general that if H and K are subgroups of a group G, and the
order of K divides the order of H, then K is a subgroup of
H. If that were true, a group could never have more than
one subgroup of a given order: if H and K were two subgroups
of the same order, the order of each would divide the order
of the other, so by your claim each would be a subgroup of
the other, and hence they'd be equal. But the Klein 4-group
is an easy example of a finite group with three different
subgroups of order 2.
I don't know what you already know at this point, so I don't
know what tools are available to you. If you know about the
connection between normal subgroups and quotient groups,
however, I suggest looking at the quotient group G/H. If K
is not a subgroup of H, then there is an x in K such that x
is not in H; if h : G --> G/H is the canonical homomorphism,
look at the order of h(x) in G/H.
Brian
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