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## Main Question or Discussion Point

I need some help with what should be a really simple group theory proof, but for some reason I'm hitting my head against a wall on this one. I seem to be missing something simple to get me to the next step.

I'm learning this stuff in Swedish, so I'm not sure about all the English words, but I'll try and make it at least clear. The problem is pretty simple. We just learned the Lagrange theorem and a few simple resulting theorems from that.

Let H and K be two normal cyclical subgroups to a finite group G. Let |H| and |K| be relatively prime. Show that:

HK = {hk | h element of H, k element of K}

is also a cyclical subgroup to G.

Showing it's a subgroup doesn't seem to be much problem, but showing it is cyclical is tripping me up. I _think_ I can do it if I show hk=kh for all h and k, but I'm not sure how to do this. I'm also not sure what the groups being normal has to do with it (if anything) since aren't all cyclical groups automatically normal (since they always commute, and all groups that commute are normal)?

I can see intuitively that the product of the generators to H and K should itself be a generator to HK, but I'm having trouble "proving" it more formally.

This feels like it should be really simple, with one little step I'm missing, but I can't see it.

Things I played around with: H and K have no undergroups in common, and I believe they have to be disjoint (one follows from the other I think, doesn't it?). I can see that the only hk that equals 1 is when they both equal one, which means that if hk commutes then (hk)^|H||K| = 1.

Anyway, any help or clues would be appreicated. Thanks.

I'm learning this stuff in Swedish, so I'm not sure about all the English words, but I'll try and make it at least clear. The problem is pretty simple. We just learned the Lagrange theorem and a few simple resulting theorems from that.

Let H and K be two normal cyclical subgroups to a finite group G. Let |H| and |K| be relatively prime. Show that:

HK = {hk | h element of H, k element of K}

is also a cyclical subgroup to G.

Showing it's a subgroup doesn't seem to be much problem, but showing it is cyclical is tripping me up. I _think_ I can do it if I show hk=kh for all h and k, but I'm not sure how to do this. I'm also not sure what the groups being normal has to do with it (if anything) since aren't all cyclical groups automatically normal (since they always commute, and all groups that commute are normal)?

I can see intuitively that the product of the generators to H and K should itself be a generator to HK, but I'm having trouble "proving" it more formally.

This feels like it should be really simple, with one little step I'm missing, but I can't see it.

Things I played around with: H and K have no undergroups in common, and I believe they have to be disjoint (one follows from the other I think, doesn't it?). I can see that the only hk that equals 1 is when they both equal one, which means that if hk commutes then (hk)^|H||K| = 1.

Anyway, any help or clues would be appreicated. Thanks.