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

1. ## 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 $\displaystyle H \cup K$ is a subgroup of G.

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I had the set of generators 0 through 17 for $\displaystyle \mathbb{Z}_{18}$ sitting in front of me from a previous problem.

I saw that

$\displaystyle <2> = \left\{0,2,4,6,8,10,12,14,16\right\}$,
$\displaystyle <9> = \left\{0,9\right\}$.

So I said $\displaystyle <2>=H$ and $\displaystyle <9>=K$.

Therefore, $\displaystyle H \cup K$ would be

$\displaystyle \left\{0,2,4,6,8,9,10,12,14,16\right\}$

But this isn't closed under the group operation. $\displaystyle 9+2 \hspace{1mm} mod \hspace{1mm} 18=11$ and $\displaystyle 11 \notin H\cupK$

So am I correct in saying that $\displaystyle H \cup K$ 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?

2. ## Re: Prove or disprove, if H,K<G => HuK<G

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).

3. ## Re: Prove or disprove, if H,K<G => HuK<G

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).

4. ## Re: Prove or disprove, if H,K<G => HuK<G

Although there is nothing wrong with Deveno's answer, it would, perhaps, be easier to see in a finite example. So, notice that $\displaystyle |A\cup B|=|A|+|B|-|A\cap B|$, 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 $\displaystyle A\cap B=1$ then you have that $\displaystyle |A|$, $\displaystyle |B|$ and $\displaystyle |A|+|B|-1$ 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 $\displaystyle pq$ will work with $\displaystyle p, q$ primes (as these must contain subgroups of order $\displaystyle p$ and of order $\displaystyle q$ which are cyclic and so intersect trivially) but $\displaystyle p+q-1$ does not divide $\displaystyle pq$ (why?).

5. ## Re: Prove or disprove, if H,K<G => HuK<G

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.

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# show that if H and K are subgroups of G then HUK is not a subgroup of G

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