Thread: Prove or disprove, if H,K<G => HuK<G

Problem: Prove or provide a counter example.
If H and K are subgroups of a group G, then is a subgroup of G.

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I had the set of generators 0 through 17 for sitting in front of me from a previous problem.

I saw that

,
.

So I said and .

Therefore, would be



But this isn't closed under the group operation. and

So am I correct in saying that will not always be a subgroup of G?

Even if I'm correct, I'm a little concerned. If I had not had the generator list in front of me, I wouldn't have immediately spotted the counterexample. Is there any mathematically concrete way to go about this?

one counter-example is all it takes to disprove something. sometimes HUK is a subgroup, for example, this will always be true if one subgroup is contained in the other.

with sets, HUK is the set generated by H and K. it is the smallest subset of some set that contains both H and K (the "universe set") that contains both H and K.

by contrast, the smallest group containing both H and K is <HUK>, and it is easy to see that while for a given set S, <S> contains S, it need not equal S.

Z is a perfectly good place to observe this phenomenon.

we know that every subgroup of Z is cyclic, and therefore it has a generator. pick two positive numbers ≠1 in Z that are relatively prime, say k and m.

let's see if <k> U <m> is a subgroup. since k,m are relatively prime, we can find integers s,t such that sk + tm = 1.

now sk is in <k>, therefore in <k> U <m>, and tm is in <m> therefore in <k> U <m>, so if <k> U <m> is a subgroup,

sk + tm = 1 is in <k> U <m>, so all of Z is in <k> U <m>, since 1 generates Z.

now consider km + 1, which, being in Z, is certainly in <k> U <m>, so is either in <k>, or in <m>.

let's suppose it is in <k>, so we have km + 1 = ku, for some u, so k(u-m) = 1

the only value for k for which this works is 1, contradicting our choice of k.

so km +1 must be in <m>, so km + 1 = mv, for some v, so m(v-k) = 1, and then m = 1.

but we chose both k,m ≠ 1, so <k> U <m> must not be a subgroup of Z.

(you can check that if k = 2, and m = 3, that 2+3 = 5 isn't a multiple of 2 or a multiple of 3).

Thanks for the extra information, it answered my next question which was, if H and K are both contained in each other, if that would be a subgroup (after several trials, it appeared to be).

Although there is nothing wrong with Deveno's answer, it would, perhaps, be easier to see in a finite example. So, notice that , which should be covered in some course you have done before.

Now, there exist subgroups of groups which intersect trivially (take, for example, two subgroups of order 2 of the Klein 4-group). So if then you have that , and all divide the order of the group (by Lagrange).

It is not, therefore, too hard to come up with a counter-example. Basically, any group of order will work with primes (as these must contain subgroups of order and of order which are cyclic and so intersect trivially) but does not divide (why?).

and the smallest example of such a pq is 6. two groups of order 6.

the first is Z6 = {0,1,2,3,4,5}. the subgroup of order 2 is H = {0,3}, the subgroup of order 3 is K = {0,2,4}.

now H U K = {0,2,3,4} which has 4 elements (2+3-1). 4 does not divide 6.

(and one can see that 2+3 = 5 is not in H U K).

the other example is S3 = {1,(1 2),(1 3),(2 3),(1 2 3),(1 3 2)}.

for H we can choose {1,(1 2)}, for K we have only one choice, K = {1,(1 2 3),(1 3 2)}.

H U K = {1,(1 2),(1 2 3),(1 3 2)}. here again, (1 2)(1 2 3) = (2 3), which is not in H U K.

the proof that p+q-1 does not divide pq is a bit tricky. there are 2 cases:

1) p = 2. then p+q-1 = q+1, while pq = 2q. if q+1 = 2q, then q = 1, but 1 is not prime.
2) p and q are odd primes. in this case, p+q-1 is even, but pq is odd, so has no even factors.