## 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.