1. ## Proof and questions about a subgroup theorem

Here is the question:
If p > q are primes, a group of order pq has at most one subgroup of order p.

Hint: Suppose H, K are distinct subgroups of order p. Show that $\displaystyle H \cap K = <e>$ and then use $\displaystyle [H \vee K : H ] \geq [K : H \cap K ]$ to show a contradiction.
Using the hint it was very easy to show a contradiction. Briefly the result is
$\displaystyle [ H \vee K : H] = \frac{p^2 - 1}{p}$ and $\displaystyle [K : H \cap K] = p$

I have two questions. The first: Is there a specific reason that $\displaystyle H \cap K = <e>$ uses <e> rather than {e}?

The second is by way of a counter-example. I note that $\displaystyle \mathbb{Z} _2 \oplus \mathbb{Z} _2$ contains 3 subgroups of order 2. Now this doesn't directly contradict the theorem since p = q = 2 and the problem statement said p > q. However the proof by contradiction never used this fact and I can find nothing in the proof to indicate that p = 2 and |G| = 4 should be a special case?

Thanks.

-Dan

2. Originally Posted by topsquark
Here is the question:
Using the hint it was very easy to show a contradiction. Briefly the result is
$\displaystyle [ H \vee K : H] = \frac{p^2 - 1}{p}$ and $\displaystyle [K : H \cap K] = p$

I have two questions. The first: Is there a specific reason that $\displaystyle H \cap K = <e>$ uses <e> rather than {e}?

The second is by way of a counter-example. I note that $\displaystyle \mathbb{Z} _2 \oplus \mathbb{Z} _2$ contains 3 subgroups of order 2. Now this doesn't directly contradict the theorem since p = q = 2 and the problem statement said p > q. However the proof by contradiction never used this fact and I can find nothing in the proof to indicate that p = 2 and |G| = 4 should be a special case?

Thanks.

-Dan
A group of order $\displaystyle p^2$ is always abelian (why?), and so is either cyclic (and so contains only one subgroup of order p) or is $\displaystyle C_p\times C_p$, and so contains 3 subgroups of order p. So, there is nothing special about p=2; it works for all p.

However, the proof you outlined doesn't work for groups of order $\displaystyle p^2$, as, $\displaystyle |H\vee K:H|=p$ and so you don't get a contradiction (assuming, of course, that I am remembering what $\displaystyle \vee$ means correctly!).

3. <e> and {e} are often used interchangeably.

it isn't true that |H v K| = p^2 -1. in this case, |H v K| has to be less than or equal to pq < p^2 (since q < p), so |H v K| is at most p^2 - 1.

4. Originally Posted by Swlabr
A group of order $\displaystyle p^2$ is always abelian (why?)
yes, why indeed? i always thought it was because G/Z(G) couldn't be non-trival cyclic.

5. Originally Posted by Deveno
yes, why indeed? i always thought it was because G/Z(G) couldn't be non-trival cyclic.
Indeed, that is why!

6. Originally Posted by Deveno
<e> and {e} are often used interchangeably.

it isn't true that |H v K| = p^2 -1. in this case, |H v K| has to be less than or equal to pq < p^2 (since q < p), so |H v K| is at most p^2 - 1.
(sighs) I was hoping to be back before someone noticed that. I was tired yesterday.

I have constructed several examples for this theorem and I can't get any contradictions. So....the only bit I could see being wrong is how I'm forming the join.

Just to be explicit I'm going to consider the (sub)groups H and K such that H = <a> of order 2 and K = <b> of order 3.
$\displaystyle H \cup K = \{e, a, b, b^2 \} \implies H \vee K = \{e, a, b, b^2, ab, ab^2 \}$

In other words, more generally, given H = <a> and K = <b> of some finite (prime) order $\displaystyle H \vee K = \{a^mb^n | 1 \leq m \leq |H|, 1 \leq n \leq |K| \}$. This will be of cardinality |H|*|K|. Am I constructing the join correctly?

-Dan

7. Originally Posted by topsquark
In other words, more generally, given H = <a> and K = <b> of some finite (prime) order $\displaystyle H \vee K = \{a^mb^n | 1 \leq m \leq |H|, 1 \leq n \leq |K| \}$. This will be of cardinality |H|*|K|. Am I constructing the join correctly?
I don't have time to give this a proper look-over, but I cannot see any reason why your group elements have this normal form. For example, aba may never be of this form...

8. the join H v K, is the smallest subgroup of G containing both H and K. as a set, it is larger than HUK, which may not be a subgroup. since H v K is a subgroup,

by closure it has to contain the set HK, which also may, or may not be a subgroup. here is an example: let G = A4, and let H = {1, (1 2 3), (1 3 2)},

and K = {1, (1 2)(3 4)}. HK has order 6: HK = {1, (1 3 2), (1 2 3), (1 2)(3 4), (1 3 4), (2 3 4)}. note that HK is NOT a subgroup, for example,

(2 3 4)^-1 = (2 4 3) is not a member. note that we have to add at least 2 inverses to HK to make it have inverses ((1 4 3) has to be added, as well),

so that |H v K| is at least 8. but 8 does not divide 12, so |H v K| must be all of A4 (which has order 12).

the join operation doesn't work very well on groups in general, it behaves much better on abelian groups (who decompose very nicely into direct sums).

a word of advice: when testing generalizations about subgroups, look at non-abelian groups.

9. Well it seems pretty clear that I need to work with the join concept then come back. Thank you all for your assistance.

-Dan

10. the join isn't all that useful. if you want to make a "bigger" group out of smaller groups, you have two basic constructions:

H x K, the direct product. this can't always be done within a group, as it places some severe restrictions on H and K (they have to be normal subgroups).

H x| K, the semi-direct product. again, this also cannot always be done within a group, as H has to be normal.

there are groups which stubbonly resist attempts to "factor" them, because they HAVE no normal subgroups.

basically, for HK to be the kind of thing we want it to be, some sort of "set-commuting" has to happen, we really want KH = HK.

if that happens we have the multiplication analogue of the sum of vector spaces U+V.

abelian groups naturally have this "niceness", but they're too restrictive. if one of H,K is normal, we can work around this.

but even with normal subgroups of a group, there is a problem, if N,K are normal, N v K may not be (you have to take the

normal closure). groups don't have a "natural" join operation.

11. Originally Posted by Deveno
H x| K, the semi-direct product. again, this also cannot always be done within a group, as H has to be normal.
More than this - you can take the (external) direct product of two arbitrary groups, but in general you cannot take the (external) semidirect product of two groups; one has to act on the other, which doesn't always happen between groups. Some groups just can't act (non-trivially) on others! (Of course, every direct product is a semidirect product, so when I said semidirect product' above I actually meant semidirect product which isn't a direct product'.)

12. right, the only possible homomorphism, K-->Aut(H) may be trivial. but the condition that HK be a group and that only H be normal, is weaker than HK is a group and both H,K are normal. (and of course, we want H∩K = {e} in either case. splitting up a group into two factors isn't all that helpful if there's overlap).

it's fair to say that H x K is universal, but H x| K is not, correct?