## Algebra Qual Study Problems: 1

2003 # 1: Prove that a group cannot be the union of two proper subgroups.
Proof-
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.
Proof-
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 $\langle K_1 \cup K_2 \rangle$.
Example-
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.
Proof-
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 $D_4 \times \mathbb{Z}_2$.
Proof-
$D_8$ 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.
Proof-
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.
Proof-
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.
Proof-
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.
Proof-
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.
Proof-
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.