Algebra Qual Study Problems: 1

2003 # 1: Prove that a group cannot be the union of two proper subgroups.
Let H and K be two proper subgroups of a group G such that H and K are not the same, otherwise their union would have to be G but they are the same group so the union would just be one group, the entire group and thus not a proper subgroup. Suppose their union is G. Let k\in K \backslash H. Then for any h\in H we have k + h \in H or k+h\in K. In the first case we get -h + (h + k) = k \in H since inverses are in subgroups. This contradicts k being in not H. In the second case we get that (h+k)+ -k = h \in K which is a similar contradiction. Thus G is not the union of H and K.

2011 #1A (Hungerford I.2 #18): Suppose G is a group and that H_1, H_2 are two distinct subgroups of G. State and prove a sufficient condition for H_1 \cup H_2 to be a group.
In order for H_1 \cup H_2 to be a group it must be closed. If we have one element from each distinct subgroup, we need the sum to be in the union. Let h_1 \in H_1 and h_2 \in H_2. If h_1+h_2 \in H_2 then we get h_1 \in H_2. Similarly if h_1+h_2 \in H_1 we get that h_2 \in H_1. Thus the only way this can happen is if one is a subgroup of the other.

2011 1B: Give an example of a group G and 2 subgroups K_1, K_2 such that K_1 \cup K_2 is strictly contained in the subgroup \langle K_1 \cup K_2 \rangle.
Let G = S_3 and K_1 = \langle (1,3) \rangle, K_2 = \langle (123) \rangle. Then we have K_1 \cup K_2 = \{e,13,123,132\} \subset S_3 while \langle K_1 \cup K_2 \rangle = S_3

Qual 2001 #1 (Hungerford I.8 #1): Show the symmetric group S_3 is not isomorphic to a direct product of two of its proper subgroups.
Note that the order of S_3 is 6, so any subgroup must be of order 2 or 3, which are prime numbers. Any group of prime order must be cyclic, and cyclic groups are abelian. The direct product of abelian groups is abelian, but S_3 is not abelian so it cannot be ismorphic to an abelian group, and hence is not isomorphic to a direct product of proper subgroups.

Qual 2005 #3A: Prove that D_8 is not isomorphic to D_4 \times \mathbb{Z}_2.
D_8 must have an element of order 8. Recall that if (g,h) \in G \times H then |(g,h)| = lcm(|g|,|h|). If g\in D_4 and h\in \mathbb{Z}_2 then |(g,h)| = 1,2 \text{ or } 4 because the order of \mathbb{Z}_2 is 2 and the order of D_4 is 8.

Hungerford #12 Sec I.6: 12) The center of the group D_n is {e} if n is odd and \mathbb{Z}_2 if n is even.
First we need to know all the normal subgroups in D_n. If n is odd, the nontrivial subgroups are the group of rotations, and all the subgroups of the group of rotations. If n is even we have two more normal subgroups, each isomorphic to D_{\frac{n}{2}}.
Let H be a normal subgroup. We have two cases, either H consists only of powers of the element a, or there is some a^ib \in H. Suppose that H = \langle a^k \rangle for some k dividing n. Then:
a^i(a^k)^ra^{-i} = (a^k)^r \in \langle a^k \rangle
b(a^k)^rb = b^2(a^{-2kr})=((a^k)^r)^{-1} \in \langle a^k \rangle because of the relation ba^r = a^{-r}b. And finally,
a^ib(a^k)^ra^ib = a^i(a^{-kr-i)}=a^{-kr} \in \langle a^k \rangle. Thus the subgroup is normal.
Now instead suppose there exists some a^kb \in H. Then we have a^{k+2}b = aa^k(ab) = a(a^kb) a^{-1} \in H since H is normal. This gives that (a^{k+2}b)(a^kb) = a^{k+2} a^{-k} = a^2\in H. This tells us that a^2 is a generator of H.
Next, a^{k-2l}b = (a^2)^{-l}(a^kb) \in H \forall l \in \mathbb{Z}. If k= 2m is even let l = m \Rightarrow b\in H and if k = 2m+1 is odd let l = m \Rightarrow ab\in H. Therefore either b is a generator or ab is a generator. So that H = \langle a^2,b \rangle or H = \langle a^2,ab \rangle.
Now that we know all the normal subgroups we can find the center, because it must be a normal subgroup. Recall the center is the set of all elements which can commute with any other element.
First we check H = \langle a^2,b \rangle. We have ab = ba \Rightarrow a^{-1} = a \Rightarrow a^2 = e, but n is larger than 2 so this is a contradiction. Thus this is not the center.
Next we check H = \langle a^2,ab \rangle. We have that:
a^2(ab) = (ab)a^2
a^3b = a^{-1}b \Rightarrow a^3 = a^{-1} \Rightarrow a^4 = e. But for n = 4, (ab)b = b(ab) \Rightarrow a = a^{-1} \Rightarrow a^2 = e again a contradiction so this is not the center.
Lastly, assume that \langle a^k \rangle is the center. Then we have ba^k = a^kb = ba^{-k} \Rightarrow a^{2k} = e \Rightarrow 2k = n \Rightarrow k = \frac{n}{2}. Thus the center can only be \langle a^{\frac{n}{2}} \rangle which the same as \mathbb{Z}_2 but only when n is even. If n is odd then the center is just the identity.

2001 Qual #3 (Hungerford 2.4.9): Prove that if G is a group and G/C(G) is cyclic then G is abelian.
Suppose that G/C(G) is cyclic. It is a set of cosets so to be cyclic means to be generated by a coset, i.e, G/C(G) = \langle gC(G) \rangle for some element g in G. Then for two elements g_1,g_2 \in G we have that:
g_1C(G) = g^rC(G)
g_2C(G) = g^sC(G)
This gives that g_1g^{-r} = x \in C(G) and g_2g^{-s} = y \in C(G). Then finally g_1g_2 = xg^ryg^s = xyg^rg^s = yg^sxg^r = g_2g_1 so that G is abelian.

1999 Qual #3 and 1995 #5 (Hungerford 2.5.13) – Prove that a group of order p^2 is abelian.
The center of the group is known to be a subgroup and therefore [G:C(G)] must divide p^2 so that [G:C(G)] = 1,p,p^2. We can eliminate some choices though. Recall the class equation:
|G| = |C(G)| + \sum\limits_{i=1}^m [G:C_G(x_i)] where x_i are distinct conjugacy classes (x_i \in G/C(G) such that [G:C_G(x_i)] > 1 \Rightarrow p^2 = |C(G)| + \sum\limits_{i=1}^m p^{k_i} because the order of any conjugacy class must divide the order of G and so must be a power of p, up to p^2. Thus, p must also divide |C(G)| so that |C(G)| is either p or p^2. Now using Lagrange we can say:
|C(G)| = p^2 \Rightarrow  [G:C(G)] = 1 \Rightarrow G=C(G) \Rightarrow G abelian
|C(G)| = p \Rightarrow  [G:C(G)] = p \Rightarrow G/C(G) is cyclic \Rightarrow G abelian.

Alternate Proof-
Since G is a p-group, the center C(G) \neq \{e\}. By Lagrange we then have either |C(G)| = p^2 which gives the center as the entire group, thus making G abelian, or we have |C(G)| = p. If |C(G)| = p then we have that G/C(G) is cyclic being of prime order, which tells us that G is abelian by 2.4.9.

Hungerford 1.3 #8: A group that has only a finite number of subgroups must be finite.
Let a group G have a finite number of subgroups but suppose G is infinite. Let g_i \in G not the identity. Consider \langle g_i \rangle. If it is infinite then it must be cyclic and isomorphic to the integers under addition, which has infinitely many subgroups. Thus \langle g_i \rangle cannot be infinite and in fact every element of G must have finite order. Next, let g_1 \in G and again form \langle g_1 \rangle. Since G is infinite, but \langle g_1 \rangle is finite we have that G\backslash \langle g_1 \rangle must be infinite as well. Choose another g_2 \in G\backslash \langle g_1 \rangle . Form \langle g_2 \rangle. Note that \langle g_1 \rangle \neq \langle g_2 \rangle. Again we can pick g_3 \in G\backslash (\langle g_1 \rangle \cup\langle g_2 \rangle ) and form \langle g_3 \rangle. Continuing in this manner we obtain an infinite sequence of subgroups of G which are not all equal. A contradiction to the finite number of subgroups. Thus G is not finite.

Hungerford 1.4 #11: Let G be a group of order 2n, then G contains an element of order 2. If n is odd and G is abelian, there is only one element of order 2.
Let the order of G be 2n. Consider the set \{e, a_1,a_1^{-1}, a_2,a_2^{-1}, ..., a_m,a_m^{-1}\}. If there does not exist an element a_i such that a_i^{-1} = a_i then the order of this set is 2m+1, which is a contradiction so there must exist at least one element of order 2.
If G is abelian and n is odd, suppose we have two elements a,b \in G such that a^2 = e = b^2. G is abelian so that (ab)^2 = e \Rightarrow \{e,a,b,ab\} is a subgroup of G. But because n is odd, we know that 4 does not divide 2n, so this subgroup cannot exist by Lagrange. Thus there can be only one element of order 2.

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