Here's a proof to kick off my monster study sess

So I'm supposed to be studying crazy hard for this PHL final, but I'm not really feelin' it. So I'm going to spend a little time doing a proof or two for Group first, to get me motivated. Here goes:
The Fundamental Theorem of Cyclic Subgroups
Claim: Every subgruop of a cyclic group is cyclic. Moreover, if |'a'|=n, then the order of any subgroup of 'a' is a divisor of n; and, for each positive divisor k of n, the group 'a' has exactly one subgroup of order k -- namely, 'a^(n/k)'.
Pf: Let G='a' and suppose that H a subgroup of G.
To prove the first claim (that every subgroup of a cyclic group is itself cyclic), we need to show that H is cyclic. Assume that H does not equal just the identity e. I claim that H contains an element of the form a^t, where t is positive. Since G='a', every element of H has the form a^t; and, when a^t belongs to H with t<0, h="'a^m'." g="'a'," b="a^k" k="mq+r," k="a^(mq+r)="(a^mq)(a^r)-">a^r=(a^-mq)(a^k); since a^k and a^-mq are both in H, a^r is in H. However, m is the least positive integer such that a^m is in H, but r is less than m and greater than or equal to 0. So, r=0, implying that b=a^k=a^-mq=(a^m)^q which is in 'a^m'. Then, the arbitrary element b is in the cyclic subgroup of H 'a^m', so H='a^m', so H is cyclic. This proves the first claim.
To prove the second claim (if |'a'|=n, then the order of any of 'a' will be a divisor of n), suppose that |'a'|=n and H is a subgroup of 'a'. H is cyclic, so let H='a^m', where m is the least positive integer such that a^m is in H. Let b=a^mq again, such that a^mq=e. Then, e=a^mq=b=a^n, implying that mq=n. So, the order of the subgroups of a cyclic group divide the order of the group.
To prove the third and final claim (for each positive divisor k of n, the group 'a' has exactly one subgroup of order k, 'a^(n/k)'), let k be any positive divisor of n. I need to show that 'a^(n/k)' is the one and only subgroup of 'a' of order k. |'a^(n/k)'|=n/gcd(n,n/k)=k. Now, let H be any subgroup of 'a' of order k. I know that H='a^m', where m divides n. Then, m=gcd(n, m) and k=|a^m|=|a^gcd(n,m)|=n/gcd(n,m)=n/m. Thus, m=n/k, and H='a^(n/k)'.
qed

That was pretty confusing. I need to work on that one.
Ok, PHL time. I'll publish another one in a bit.