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.

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