Abstract Algebra!!!?

Question

Suppose that a cyclic group G has exactly three subgroups: G itself, {e}, and a subgroup of order 7. How many elements are in G? What can you say if 7 is replaced with p where p is a prime?

(e is the identity)

Ok, aren't e and G always subgroups of G? Is the answer supposed to be 7? I don't know how to think about this.

Answer 1

Since it's known to be a CYCLIC group, we don't have to go to the Sylow theorems or anything like that.

Let g be a generator of a cyclic group of order N. For any factorization n=ab, g^a generates a subgroup of order b.

Hence, the total number of subgroups is at least the number of factors of n. (Actually, "at least" could be replaced with "exactly", but you don't need to prove that to solve the problem.)

So you know the order of G has at most 3 factors, one of which is p. So it equals either p or p^2. And the case where it's p is obviously ruled out (then there would only be two subgroups in all).

So we're done!

Answer 2

Okay let's see what happens when we take p to be 7. You're absolutely right that and G are always subgroups of G. The problem is telling us that there exists some other subgroup (call it H) of order 7. By Lagrange's Theorem, since H is a subgroup and o(H) is 7, the order o(G) of G must be divisible by 7. So o(G) = 7*n for some integer n > 0. Let's try some n's.

n=1 doesn't work. If o(G) = 7, then G=H since there is only one group of order 7, and that would mean G only has 2 subgroups, not 3 as required.

n=2 doesn't work. That would make G cyclic of order 14, and you'd have an extra subgroup <7> of order 2. Too many subgroups.

n=3 has the same problem. Now <7> has order 3.

n=4 is a nightmare. <2> has order 14, <7> has order 4, and <14> has order 2.

So we're seeing that n must contain only factors of 7; otherwise we pick up extra subgroups of order n. We've already ruled out n=1, so let's try n=7.

n=7 implies that G is a cyclic group of order 49. By Lagrange, the only possible subgroups are , <7>, and G. We know that and G are subgroups, and clearly <7> is a subgroup of order 7. So 49 works as the order of G.

Now we need to show that n>7 doesn't work. We can still restrict n to powers of 7, so that means we're supposing that n=7^m for some integer m>2. Immediately we have a problem: <7> is a subgroup of order 7^m, since o(G)=7^(m+1). This is a nice contradiction.

So it must be that o(G) = 49. I didn't know the answer beforehand, so I'm sure you can clean up my extraneous work when you do it yourself. I think it's pretty clear that the order of G is p^2 if you replace 7 with any prime p. The argument is exactly the same as above. In fact, you could do the argument with p first, and then conclude that it's true in the particular case of 7 since 7 is a prime.

Answer 3

Yuck...I've never heard of any of this...good luck!