Math 417 Homework 4 Solutions

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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 = xⁱ 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>.
[3] Let S, S₁ and S₂ be subsets of a group G.
- (a) If S₁ Ì S₂, show that C_G(S₂) Ì C_G(S₁).
  
  If g Î C_G(S₂), then gs = sg for every element s of S₂. But S₁ Ì S₂, so gs = sg for every element s of S₁, hence g Î C_G(S₁). Thus C_G(S₂) Ì C_G(S₁).
- (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).
[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 7², 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 = <gⁱ> 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).