## Algebra Qual Study Problems: 2

Qual 1994 # 3(Hungerford 1.5.1): (a) If N is a subgroup of index 2 in a group G, then N is normal in G.
Proof-
Let $L = \{aN,bN\}$ be the set of distinct left cosets of N in G. Let $R = \{Nc,Nd\}$ be the set of distinct right cosets of N in G. N itself is always both a right and left coset of a subgroup N of G. Also, cosets partition the group so we can write:
$L = \{N,G\backslash N\} = R$
Thus, every left coset of N in G is also a right coset of N in G so that N is normal in G.

b) Give an example of a subgroup $H \leq G$ such that $[G:H] = 3$ and $H \ntriangleleft G$.
Example- Let $G = S_3$ and $H = \langle (12) \rangle$. Since $|S_3| = 6$ and $|H| = 2 \Rightarrow [G:H] = 3$ by Lagrange. But we have that $(23)(12)(23) = (2)(31) \notin H$. So that H is not normal in G.

201A Midterm (2010) #5: What is the natural homomorphism of groups $\mathbb{Z} \rightarrow \mathbb{Z}_6$? What is the kernal of this map? Use the 1st homomorpism theorem to conclude that there is a homomorphism map from $\mathbb{Z}_{18} \rightarrow \mathbb{Z}_6$. Prove that the kernel of this homomorphism is isomorphic to $\mathbb{Z}_3$. Is it true that $\mathbb{Z}_{18} \cong \mathbb{Z}_6 \times \mathbb{Z}_3$?
Proof-
Let $\phi : \mathbb{Z} \rightarrow \mathbb{Z}_6$ by $m \mapsto m \pmod{6}$ is a natural homomorphism with $Ker\phi = 6\mathbb{Z}$. Since $18\mathbb{Z}$ is contained in the Kernel of the map we have by the first homomorphism theorem that there is homomorphism map $f: \mathbb{Z}_{18} \rightarrow \mathbb{Z}_6$ such that $\phi = f \circ \pi$ where $\pi$ is the natural projection $\pi : \mathbb{Z} \rightarrow \mathbb{Z}_{18}$. Now let $[x] \in Ker(f)$. Then we have $\phi (x) = f(\pi (x)) = f([x]) = 0$. This says that $x \in 6\mathbb{Z}$ so that $6|x \Rightarrow 6 | [x] \Rightarrow [x] = 0,6,12$. Now, these elements are all contained in $Ker(f)$ so that $Ker(f) = \{0,6,12\} \cong \mathbb{Z}_3$. Lastly, $\mathbb{Z}_{18} \ncong \mathbb{Z}_6 \times \mathbb{Z}_3$ because $\mathbb{Z}_{18}$ is cyclic while $\mathbb{Z}_6 \times \mathbb{Z}_3$ is not since 6 and 3 are not relatively prime.

Hungerford 1.8 #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.

Qual 2013 #1 (Hungerford 1.8 #12): A) Suppose that H is a normal subgroup of G and that $G \cong H \times K$. Prove that K is isomorphic to a normal subgroup of G.
Proof-
Let $\phi$ be the isomorphism given. $\phi : H \times K \rightarrow G$. Let $\pi_1 : H\times K \rightarrow H$ be the canonical projection so that $Ker(\pi_1) = K$ and recall that the kernel is always a normal subgroup. To finish the proof we need a side lemma. If $f: A \rightarrow B \rightarrow 0$ is a surjective group hom, with N normal in A, then $f(N)$ is normal in B. Proof- $bf(n)b^{-1} = f(a)f(n)f(a^{-1}) = f(ana^{-1}) \in f(N)$ so that $f(N)$ is normal in B.
Now, since $\phi$ is surjective and since $Ker(\pi_1 ) \triangleleft H \times K$ we get that $\phi (Ker(\pi_1)) \triangleleft G$ by the lemma. But $\phi (Ker(\pi_1)) = \phi (K) \cong K$.

B) Suppose G, H, K are as in part A). Prove that any homomorphism of groups $\phi : H \rightarrow H$ gives rise to a canonical homomorphism from $\widetilde{\phi} : G \rightarrow G$. Give an example to show that $\phi$ an isomorphism need not imply $\widetilde{\phi}$ is an isomorphism.
Proof-
Let $\phi : H \rightarrow H$ be a group homomorphism. Let $g\in G$ and $(h,k) \in H \times K$ be the unique element that is associated to g under the isomorphism, $G \cong H \times K$. Define $\widetilde{\phi} : G \rightarrow G$ by $\widetilde{\phi}(g) = \phi (h)$. Then clearly $\widetilde{\phi}$ is a group homomorphism. Next, Let $G = D_6, H = \langle a^3 \rangle, K = S_3$. Then we have $D_6 \cong \langle a^3 \rangle \times S_3$. The only isomorphism possible is $\langle a^3 \rangle \rightarrow \langle a^3 \rangle$, which is the identity. So then $Im(\widetilde{\phi} ) \cong \langle a^3 \rangle \neq D_6$. So that $\widetilde{\phi}$ is not an isomorphism.

2011 Qual #2 (Hungerford 1.8.9): Suppose that G is a group and H, K are subgroups of G such that G is the internal direct product of H and K. Prove that $G/H \cong K$. Deduce that there does not exist a subgroup H of $S_5$, such that $S_5 \cong H \times A_5$.
Proof-
First, let $G = H \times K$ and consider the projection map $\pi : H \times K \rightarrow K$. Note that this map is surjective since K is a subgroup of G. If $x = hk \in Ker(\pi )$ then $\pi (hk) = e$ which says that $k = e \Rightarrow x = h \in H$. The implications can be reversed so that $Ker( \pi ) = H$. Then we have by the first isomorphism theorem that $(H \times K ) / H \cong K \Rightarrow G/H \cong K$.
Suppose now that there is a subgroup H of $S_5$ such that $S_5 \cong A_5 \times H.$ Then we must have that $S_5 / A_5 \cong H$ but we have that $S_5/A_5 \cong \mathbb{Z}_2$ so that $H = \mathbb{Z}_2$. Now consider the element $(12345) = (15)(14)(13)(12) \in A_5$. This gives that the element $((12345),1)$ is isomorphic to some element in $S_5$ of order 10, call it $\sigma$. There is no single cycle of order 10 so that $\sigma = \sigma_1 \sigma_2$, where $|\sigma_1| = 5, |\sigma_2| = 2$. But because $\sigma_1$ has order 5, it cannot be disjoint from $\sigma_2$ which is a contradiction.

Qual 2011 #4, Qual 2001 #4, Qual 2003 #4 (Hungerford 2.1.4): Prove that free abelian group is a free group iff it’s cyclic.
Proof-
$(\Rightarrow )$ Suppose that G is a free abelian group which is also a free group on the set X. Suppose that $|X| \geq 2$. Then there exist distinct $x,y \in X \Rightarrow xyx^{1}y^{-1} \neq q$. This gives that $xy \neq yx$ which is a contradiction to the abelian assumption. Therefore $|X| = 1 \Rightarrow$ G is cyclic.
$(\Leftarrow )$ Suppose that G is a free abelian group which is cyclic. Then $G = \langle a \rangle$ for some $a\in G$. Then each element of G has the form $a^n$ for some integer n. Define $f: G \rightarrow \mathbb{Z}$ by $a^n\rightarrow n$ is a group isomorphism so that G is free since $\mathbb{Z}$ is free.
Counterexample: $Z_m$ is cyclic but not free because it has finite order, and all nontrivial free groups consist of elements of infinite order.

Hungerford 2.1.10b: Show that $\mathbb{Q}$ is not free abelian.
Proof-
Suppose that $\mathbb{Q}$ was free so that it would have a nonempty basis, X. If $|X| = 1$ then we have that $\mathbb{Q}$ is cyclic. Suppose it were generated by one element x, then $\frac{x+2x}{2} \notin \mathbb{Q}$ which is a contradiction. So instead suppose that $|X| \geq 2$. Then let $\frac{a}{b}, \frac{c}{d} \in \mathbb{Q}$. We have that $bd\frac{a}{b} + (-ad)\frac{c}{d} = 0$ so that the arbitrary elements are not linearly independent. Thus $|X| \ngeq 2$ and so the rationals do not have a basis and are not a free group.