## Algebra Solutions: Hungerford I.8

1) (Qual 2001 #1) $S_3$ is not the direct product of any family of its proper subgroups. The same is true of $\mathbb{Z}_{p^n}$ and $\mathbb{Z}$
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.
Now, let p be prime and $n \in \mathbb{N}$. We have $\mathbb{Z}_{p^n}$ is cyclic of order $p^n$. Thus any proper subgroup of $\mathbb{Z}_{p^n}$ must be of the form $\mathbb{Z}_{p^k}$ with $1\leq k . Suppose that:
$\mathbb{Z}_{p^n} \cong \mathbb{Z}_{p^{k_1}} \times \cdots \times \mathbb{Z}_{p^{k_r}}$. Consider the subgroups of this direct product:
$\langle p^{k_1-1} \rangle \times \{0\} \times \cdots \times \{0\}$ and
$\{0\} \times \cdots \times \{0\} \times \langle p^{k_r-1} \rangle$
Both have order p. This is because $p \cdot p^{k_1-1} = p^{k_1} = 0$ in $\mathbb{Z}_{p^{k_1}}$ and $p \cdot p^{k_r-1} = p^{k_r} = 0$ in $\mathbb{Z}_{p^{k_r}}$. Thus we have distinct subgroups of $\mathbb{Z}_{p^{k_1}} \times \cdots \times \mathbb{Z}_{p^{k_r}}$ which is impossible if $\mathbb{Z}_{p^{k_1}} \times \cdots \times \mathbb{Z}_{p^{k_r}}$ is cyclic because subgroups of cyclic groups are unique for each order. So we have a contradiction, thus $\mathbb{Z}_{p^n} \ncong \mathbb{Z}_{p^{k_1}} \times \cdots \times \mathbb{Z}_{p^{k_r}}$

4). Give an example to show that the weak direct product is not a coproduct in the category of all groups.
Proof-
Consider the weak direct product $\mathbb{Z}_2 \times \mathbb{Z}_3$ with the canonical injections $i_1: \mathbb{Z}_2 \rightarrow \mathbb{Z}_2 \times \mathbb{Z}_3$ and $i_2: \mathbb{Z}_3 \rightarrow \mathbb{Z}_2 \times \mathbb{Z}_3$. Next define the group homomorphisms, $\phi_1: \mathbb{Z}_2 \rightarrow S_3$ by $1\mapsto (12)$ and $\phi_2: \mathbb{Z}_3 \rightarrow S_3$ by $1\mapsto (123)$. Now then we have a family of groups and maps into the weak direct product along with maps into a different group. In order for the weak direct product to be a coproduct we would now need a map from the weak direct product into the different group such that the diagram commutes. So suppose there exists $\phi : \mathbb{Z}_2 \times \mathbb{Z}_3 \rightarrow S_3$ such that $\phi \circ i_j = \phi_j$. Then we have that $(12), (123) \in Im(\phi )$. But these elements generate all of $S_3$ so that $S_3 \cong Im(\phi )$. But $Im(\phi ) = \mathbb{Z}_2 \times \mathbb{Z}_3$ which is abelian, while $S_3$ is not abelian. Therefore such a map does not exist.

5) Let G, H be finite cyclic groups. Then $G\times H$ is cyclic iff $gcd(|G|,|H|) =1$
Proof-
First we need to prove another fact. The order of an element $(g,h) \in G \times H$ is $l = lcm(|g|,|h|)$ Note that $(g,h)^l = (g^l,h^l) = (e_G,e_H)$. Suppose there is some $m s. t. $(g,h)^m = (e_G,e_H)$. Then $g^m = e_G$ and $h^m = e_H$, so the order of g divides m and the order of h divides m. This gives m as a common multiple of the orders of g and of h, but l is the least common multiple so that $l \leq m$ which is a contradiction. Thus $|(g,h)| = l$. Now we can do the proof.
$(\Rightarrow )$ Assume that $G\times H$ is cyclic so that $G\times H = \langle (g,h) \rangle$ for some $g\in G, h\in H$. Let $gcd(|G|,|H|) = d$. Then we have:
$(g,h)^{\frac{|G||H|}{d}} = (g^{\frac{|G||H|}{d}},h^{\frac{|G||H|}{d}}) = ((g^{|G|})^{\frac{|H|}{d}},(h^{|H|})^{\frac{|G|}{d}}) = (e,e)$. This tells us that $|(g,h)|$ divides $\frac{|G||H|}{d}$ but $|(g,h)| = |G \times H | = |G||H|$ which means that d = 1.
$(\Leftarrow )$ Assume now that $gcd(|G|,|H|) = 1$. We know that $G = \langle g \rangle$ and $H = \langle h \rangle$ for some g, h. By the fact above we have
$| \langle (g,h) \rangle | = |(g,h)| = lcm(|g|,|h|) = lcm(|G|,|H|) = |G||H| = |G \times H|$ Therefore (g,h) generates the entire group so that $G\times H$ is cyclic.

6) Every finitely generated abelian group $G \neq \langle e\rangle$ in which every non identity element has prime order p is isomorphic to $\mathbb{Z}_p \bigoplus \cdots \bigoplus \mathbb{Z}_p$ (n summands) for some natural number n.
Proof-
Suppose G is finitely generated abelian and let $X = \{ x_1,...,x_n\}$ be a minimal generating set of G. Since G is abelian all subgroups are normal so that $\langle x_i \rangle \triangleleft G$. Also $G = \langle \cup_{i=1}^n \langle x_i \rangle \rangle$. We need only show that $\langle x_j \rangle \cap \langle \cup_{i \neq j} \langle x_i \rangle \rangle = \{e\}$. Let $a\in \langle x_j \rangle \cap \langle \cup_{i \neq j} \langle x_i \rangle \rangle$ and suppose $a\neq e$. Then it must be the case that $\langle a \rangle \leq \langle x_j \rangle$, but every element has order p by assumption so that $p = |a| = |x_j|$ which gives that $\langle x_j \rangle = \langle a \rangle \leq \langle \cup_{i\neq j} \langle x_j \rangle \rangle$. This tells us that anything in $\langle x_j \rangle$ is also in $\langle \cup_{i\neq j} \langle x_j \rangle \rangle$ which gives that $x_j$ is redundant so G can be generated without the element $x_j$. This contradicts the minimality of the generating set. Thus $a = e$. Now from theorem 8.6 we can say $G \cong \langle x_1 \rangle \times \cdots \times \langle x_n \rangle$ But each $\langle x_i \rangle$ is cyclic of order p and hence isomorphic to $\mathbb{Z}_p$. Cyclic groups are abelian so we can rewrite this as $G \cong \mathbb{Z}_p \bigoplus \cdots = \bigoplus \mathbb{Z}_p$ (n summands)

9) If a group G is the (internal) direct product of its subgroups H, K then $H \cong G/K$ and $G/H \cong K$.
Proof-
Let $G = H \times K$ and consider the canonical projection maps:
$\pi_H: H \times K \rightarrow H$ and $\pi_K: H \times K \rightarrow K$ by $\pi_H(h,k) = h$ and $\pi_K(h,k) =k$. Then note that $\pi_H(x,y) = e$ iff $y = e$ so that $Ker\pi_H = K \times \{e\}$. Clearly also $\pi_H$ is surjective. Then by the first isomorphism theorem we have:
$(H\times K)/Ker\pi_H \cong H \Rightarrow (H\times K)/(K \times \{e\}) \cong H \Rightarrow (H\times K)/K \cong H \Rightarrow G/K \cong H$ The other isomorphism is shown in the exact same manner.