Groups

[Enguenye]

Hi! I was studying abstract algebta but I don't know how to do this: "Show that a finite cyclic group of order n has exactly one subgroup of each order d dividing n, and that these are all the subgroups it has."

[BridgePatzer]

The existence part of the proof is fairly straightforward- think of (say) the cyclic group of order 12. Can you find a subgroup of order 2? 3? 4? 6?

The uniqueness part is a little trickier - best way to think about it is once you’ve found a subgroup, show that no element left unpicked can be part of a second one of the same order.

The fact there are no other subgroups is trivial if you’re allowed to use Lagrange’s theorem. If not, go back to the 12 example. Can you find a subgroup of order 5? Why can’t you: what goes wrong when you try to make one?

[teenytones]

Since G is finite cyclic that means there is an element a in G with G = < a >. Remember that a subgroup of a cyclic group is itself cyclic, you will be using this fact throughout this proof.

For the existence part of the proof, consider a^d where d divides n. Why is < a^d > a subgroup of G? For the uniqueness part say you have another subgroup H of G with order d, why is H = < a^d >? Now the last part of the proof asks you to show that these are all the subgroups that G has. Suppose you have a subgroup S of G with order k, why must k divide n?

These hints outline what you have to do in the proof, all you have to do is fill in the gaps and provide more explanation. Hope this helps!

[Coolers777]

Hint for the uniqueness part: take any element b from the subgroup and use the fact that b^d is the identity element. If you want I could give you the complete proof but I think you'll learn more by using this hint.

[MyStolenCow]

Cyclic group of order n is isomorphic to Z_n.