Algebra Solutions: Hungerford I.5

1) 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.

2) If \{ N_i\}_{i\in I} is a family of normal subgroups of a group G, then \cap_{i\in I} N_i is a normal subgroup of G.
Proof-
Let g\in G and n \in \cap_{i\in I} This says that n\in N_i \forall i \in I. Since each N_i is normal we have that gng^{-1} \in N_i, \forall i \in I. Thus gng^{-1} \in \cap_{i\in I} N_i so that \cap_{i\in I} N_i is a normal subgroup of G.

3) Let N be a subgroup of a group G. N is normal in G iff congruence modulo N is a congruence relation.
Proof-
(\Rightarrow ) We know that congruence modulo N is an equivalence relation on G by previous exercise so we only need to show if a_1 \sim b_1 \text{ and } a_2 \sim b_2 \text{ then } a_1a_2 \sim b_1b_2. Assuming the hypothesis we have that a_1 \equiv b_1 \pmod{N} and a_2 \equiv b_2 \pmod{N}. This tells us that a_1b_{1}^{-1} \in N and a_2b_{2}^{-1} \in N. Then we have a_1b_{1}^{-1} =n_1 and a_2b_{2}^{-1} =n_2 for some n_1,n_2 \in N. Then note:
a_1a_2b_2^{-1}b_1^{-1} = n_1b_1n_2b_2b_2^{-1}b_1^{-1}= n_1b_1n_2b_1^{-1}. We have that b_1n_2b_1^{-1} \in N because N is normal in G. Thus n_1b_1n_2b_1^{-1} \in N as well by closure of N. Therefore we have:
a_1a_2(b_1b_2)^{-1} \in N so that a_1a_2 \sim b_1b_2
(\Leftarrow ) Let a\in G, n\in N. Since N is a subgroup of G, we can write n = ne^{-1} \in N so that n \equiv e \pmod{N}. Also a\equiv a \pmod{N}. Thus we can use the congruence relation property to say that:
an \equiv ae =a\pmod{N}. This tells us that ana^{-1} \in N which gives that N is normal in G.

7) Let G be a finite group and H a subgroup of G with order n. If H is the only subgroup of G of order n, then H is normal in G.
Proof-
We know from previous exercise that H \cong aHa^{-1} and so n = |H| = |aHa^{-1}|. The uniqueness of H implies that H = |aHa^{-1} which by definition gives H as normal in G.

9)(a) If G is a group then the center of G is a normal subgroup of G.
Proof-
Note that the center of G is the set C(G) = \{ c\in G \mid ca = ac, \forall a \in G\}. This set is already a subgroup of G. Let g\in G, c\in C(G). Then gcg^{-1} = gg^{-1}c = c \in C(G) so the center is normal in G

12) If H is a normal subgroup of a group G such that H and G/H are finitely generated, then so is G.
Proof-
Let H = \langle x_1,...,x_n \rangle and G/H =  \langle y_1H,...,y_mH \rangle. Claim that G = \langle x_1,...,x_n,y_1,...,y_m \rangle. Note first that \langle x_1,...,x_n,y_1,...,y_m \rangle \subset G by closure of the group G. Let g\in G. Then gH\in G/H so that we can write gH = (g_1H)^{n_1}\cdot \cdot \cdot (g_kH)^{n_k}, where each g_i = y_j for some 1\leq j \leq m and n_i \in \mathbb{Z}. Since H is normal in G, we have G/H is a group under multiplication so we can change gH = (g_1H)^{n_1}\cdot \cdot \cdot (g_kH)^{n_k} = g_1^{n_1} \cdot \cdot \cdot g_k^{n_k}H. By definition of cosets this gives g^{-1}(g_1^{n_1} \cdot \cdot \cdot g_k^{n_k}) \in H =  \langle x_1,...,x_n \rangle. Thus g^{-1}(g_1^{n_1} \cdot \cdot \cdot g_k^{n_k}) = h_1^{m_1} \cdot \cdot \cdot h_l^{m_l}, where each h_i = x_j for some 1 \leq j \leq n and m_i \in \mathbb{Z}. It then follows that g = g_1^{n_1} \cdot \cdot \cdot g_k^{n_k}(h_1^{m_1} \cdot \cdot \cdot h_l^{m_l})^{-1} \in \langle x_1,...,x_n,y_1,...,y_m \rangle

13) (a) Let H, K be normal in G. Show that H \vee K is normal in G.
Proof-
Let g\in G, n \in H\vee K. By definition of the join of H and K we can write n = a_1^{n_1}a_2^{n_2}\cdot \cdot \cdot a_k^{n_k} where a_i \in H \cup K, and n_i \in \mathbb{Z}. Then:
gng^{-1}= ga_1^{n_1}a_2^{n_2}\cdot \cdot \cdot a_k^{n_k}g^{-1} = ga_1^{n_1}g^{-1}ga_2^{n_2}g^{-1}g\cdot \cdot \cdot g^{-1}ga_k^{n_k}g^{-1}. Now if a particular a_i^{n_i} \in H or a_i^{n_i} \in K we have that ga_i^{n_i}g^{-1} \in H or ga_i^{n_i}g^{-1} \in K since H and K are both normal in G. This each ga_i^{n_i}g^{-1} \in H \cup K for any i which gives that gng^{-1}= ga_1^{n_1}g^{-1}ga_2^{n_2}g^{-1}g\cdot \cdot \cdot g^{-1}ga_k^{n_k}g^{-1} \in \langle H \cup K \rangle = H \vee K. Thus the join of H and K is normal in G.

14) If N_1 \triangleleft G_1 and N_2 \triangleleft G_2 then (N_1 \times N_2) \triangleleft (G_1 \times G_2) and (G_1 \times G_2) / (N_1 \times N_2) \cong (G_1/N_1) \times (G_2/N_2).
Proof-
Let (g_1,g_2) \in G_1 \times G_2 \text{ and } (n_1,n_2) \in N_1 \times N_2. Then (g_1,g_2)(n_1,n_2)(g_1,g_2)^{-1} = (g_1n_1g_1^{-1}, g_2n_2g_2^{-1}) \in N_1 \times N_2 because N_1 \triangleleft G_1 and N_2 \triangleleft G_2. Thus (N_1 \times N_2) \triangleleft (G_1 \times G_2).
Now we define f:G_1\times G_2 \rightarrow (G_1/N_1) \times (G_2/N_2) by f(g_1,g_2) = (g_1N_1,g_2N_2). This function is clearly well defined and surjective. Note that:
f((g_1,g_2)(g_1'g_2')) = f(g_1g_1',g_2g_2') = (g_1g_1'N_1,g_2g_2'N_2) = (g_1N_1g_1'N_1,g_2N_2g_2'N_2) = (g_1N_1, g_2N_2) (g_1'N_1,g_2'N_2) = f(g_1,g_2) f(g_1',g_2') so that f is a homomorphism. Thus f is an epimorphism. Lastly, we have Kerf = N_1\times N_2 because (g_1,g_2) \in Kerf whenever f(g_1,g_2) = (N_1,N_2) which can only happen when g_1 \in N_1 and g_2 \in N_2. So by the first isomorphism theorem we have the conclusion:
(G_1 \times G_2) / (N_1 \times N_2) \cong (G_1/N_1) \times (G_2/N_2).

15) Let N \triangleleft G and K \triangleleft G. If N\cap K = \{e\} and N \vee K = G then G/N \cong K
Proof-
Let \pi : G \rightarrow G/N be the canonical epimorphism and consider \pi \mid_K : K \rightarrow G/N. The restriction of a group homomorphism to a group is still a group homomorphism. Let a \in Ker\pi \mid_K. Then a\in K and a\in Ker \pi = N. So then a\in N \cap K = \{e\} so that the restriction is injective. Next, suppose that gN \in G/N. Then g\in G = N \vee K so that g = n_1^{m_1}k_1^{l_1} \cdots n_p^{m_p}k_q^{l_q}. But since K and N are both normal in G we can write any kn = n'k' for k,k' \in K, n,n' \in N. Thus g can be written g = k_0n_0, k_0\in K, n_0 \in N. Then \pi \mid_K (k_0) = k_0N = gn_0^{-1}N = gN so that the map restricted to K is surjective. Thus the map is an isomorphism so we are done.

19) If N \triangleleft G, [G:N] is finite, H \leq G, |H| is finite, and gcd([G:N],|H|) = 1, then H \leq N
Proof-
Let h\in H. Since H is finite so is |h|. Consider hN \in G/N. Note that (hN)^{|h|} = h^{|h|}N = eN = N \Rightarrow |hN| \mid |h|. By Lagrange we also have that |h| \mid |H|. Thus |hN| \mid |H|. Also, because hN \in G/N the order of hN must divide |G/N| = [G:N]. Therefore, |hN| is a common divisor of [G:N] \text{ and } |H|, but we know the gcd of these two numbers is 1 so that |hN| = 1, which tells us that hN is the identity element in G/N, i.e. hN = N so that h\in N. Thus H is a subgroup of N.

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