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- [1] Let H and K be subgroups of a group G. Prove that HÇK
is a subgroup of G.
[Note: the symbol Ç should be an intersection symbol.]

First, note that HÇK is nonempty since e is in both H and K. And if g and h are in both H and K, so is gh^{-1}. Thus HÇK is a subgroup by Theorem 3.1.

- [2] Let g be an element of a group G.
- (a) If x Î
<g>, show that
<x>
Ì
<g> and hence that
|x| £ |g|.
[Note: the symbol Î should be "is an element of" symbol;
the symbols < and >
should be the symbols showing that <g>
is the cyclic group generated by g;
the symbol £ should be a "less than or equals" symbol; and
the symbol Ì should be "is a subset of" symbol.]

If y Î <x>, then y = x^{i}for some integer i. But since x Î <g>, we know x = g^{j}for some integer j. Thus y = (g^{j})^{i}= g^{ij}, so y Î <g>. Thus <x> Ì <g>. But now |x| = |<x>| £ |<g>| = |g|.

- (b) Use (a) to conclude that <g
^{-1}> = <g>.

Since g^{-1}Î <g>, we see by (a) that < g^{-1}> Ì <g>. But likewise, g Î <g^{-1}>, so < g > Ì <g^{-1}>. Thus < g > = <g^{-1}>.

- (a) If x Î
<g>, show that
<x>
Ì
<g> and hence that
|x| £ |g|.
[Note: the symbol Î should be "is an element of" symbol;
the symbols < and >
should be the symbols showing that <g>
is the cyclic group generated by g;
the symbol £ should be a "less than or equals" symbol; and
the symbol Ì should be "is a subset of" symbol.]
- [3] Let S, S
_{1}and S_{2}be subsets of a group G.- (a) If S
_{1}Ì S_{2}, show that C_{G}(S_{2}) Ì C_{G}(S_{1}).

If g Î C_{G}(S_{2}), then gs = sg for every element s of S_{2}. But S_{1}Ì S_{2}, so gs = sg for every element s of S_{1}, hence g Î C_{G}(S_{1}). Thus C_{G}(S_{2}) Ì C_{G}(S_{1}).

- (b) Show that C
_{G}(S) = Ç_{s ÎS}C_{G}(s).

Since {s} Ì S for every s Î S, we know by part (a) that C_{G}(S) Ì C_{G}(s) for every s Î S, and hence that C_{G}(S) Ì Ç_{s ÎS}C_{G}(s). So now consider an element x of Ç_{s ÎS}C_{G}(s). Since x is in C_{G}(s) for every s in S, we see xs = sx for every s in S, so x Î C_{G}(S). Thus Ç_{s ÎS}C_{G}(s) Ì C_{G}(S), so C_{G}(S) = Ç_{s ÎS}C_{G}(s).

- (a) If S
- [4] Let G be a group which has exactly three different subgroups, including
a proper subgroup H of order 7. Show that G is cyclic, and determine |G|.

Since H is a proper subgroup, we can pick an element g of G that is not in H. But the only subgroups of G are {e}, H and G, and g is not in {e} or H, so <g> must be G itself. Thus G is cyclic. (It is also finite, since every nontrivial subgroup of an infinite cyclic group is infinite.) Now 7 divides |G| by Theorem 4.3, so we can write |G| = 7k for some positive integer k. Since G has exactly 3 subgroups, |G| can have only three positive divisors. Two of them are 1 and 7; if k = 1, these are the only two, but that would mean that G has only two subgroups. Thus k is bigger than 1. But if k is bigger than 1 but not 7, then |G| has at least four divisors: 1, 7, k and 7k. This would mean that G has at least four subgroups. Thus k must be 7 and the prime factorization of |G| must be 7^{2}, so |G| = 49.

- [5] Let G be a cyclic group of order n. Let D and M be subgroups of G
of orders d and m respectively.
Determine the order of D Ç M,
in terms of d, m and n.

Let h denote |D Ç M|. Let g denote the greatest common divisor, gcd(d,m). Since D Ç M is a subgroup of D, we know by Theorem 4.3 that h divides d. Likewise, h divides m. Thus h is a common factor of d and m, hence less than or equal to the greatest common factor; i.e., h £ g. But G has a unique subgroup, call it N, of order g. Since g divides D and M, both D and M also have subgroups of order g. Since D and M are in G, these order g subgroups are in G and hence must be N. Thus N is in D Ç M, so g = |N| £ |D Ç M| = h. Thus h = g; i.e., |D Ç M| = gcd(d,m).

Here's an alternative approach. If D or M is the trivial subgroup, then so is D Ç M, in which case |D Ç M| = 1. Now assume that neither D nor M is trivial. Since G is cyclic, we know G = <g> for some element g Î G, and thus D = <g^{i}> and M = <g^{j}>, where we may assume i = n/d and j = n/m by Theorem 4.3. But we also have D Ç M = <g^{k}>, where k is the least positive integer such that g^{k}is in D Ç M. Since g^{k}is in D we see k is a multiple of i, and since g^{k}is in M, k is a multiple of j, and since k is the least such positive integer, we see that k is the least common multiple lcm(i, j) of i = n/d and j = n/m. Thus |D Ç M| = n/k = n/lcm(n/d, n/m). (Note that this formula works also when either d or m is 1, even though we treated that case separately.)

It is interesting to note that the two approaches together thus show that gcd(d, m) = n/lcm(n/d, n/m).