Math Forum / Mathematics / Undergraduate Math / April 2008

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?

Yes.

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).

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.

On Tue, 15 Apr 2008 12:52:48 EDT, allpro
<nowimpsbball@yahoo.com> wrote in
<news:3753386.1208278400069.JavaMail.jakarta@nitrogen.mathforum.org>
in alt.math.undergrad:

Prove that *what* are finite?  You know that G, H, and K are
finite, because you know their orders.

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.

No, n >= k.  But it's still true that p^k | p^n.

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