## Algebra Solutions: Hungerford II.4

1) Let G be a group and A a normal abelian subgroup. Show that $G/A$ operates on A by conjugation and obtain a homomorphism $G/A \rightarrow AutA.$
Proof-
Define $G/A \times A \rightarrow A$ by $(gA,a) \mapsto gag^{-1}$ and show it is a well defined map. Let $(g_1A,a_1) = (g_2A,a_2)$. Then $g_1A = g_2A$ and $a_1 =a_2$. This gives that $g_1^{-1}g_2 \in A$ so that $g_1^{-1}g_2 = a\in A$ for some element in A. Then $g_1^{-1} = ag_2^{-1}$ and $g_1 = g_2 a^{-1}$. Then finally we have:
$g_1a_1g_1^{-1} = g_2a^{-1}a_1ag_2^{-1} = g_2a_1g_2^{-1} = g_2a_2g_2^{-1}$ because A is abelian so elements in A can commute with each other so the map is well defined. Now we show it is a group action.
Note that $(eA,a) \mapsto eae^{-1} = a$ and that:
$((g_1Ag_2A),a) \mapsto g_1g_2a(g_1g_2)^{-1} = g_1g_2ag_2^{-1}g_1^{-1} = (g_1A, g_2ag_2^{-1}) = (g_1A, (g_2A,a))$ thus the map is a group action. This action induces a group homomorphism $G/A \rightarrow AutA$ defined by $gA \mapsto \tau_{gA}$ where $\tau_{g A} :A \rightarrow A$ is given by $\tau_{g A} (a) = (gA,a) = gag^{-1}$

2) If H, K are subgroups of G such that $H \triangleleft K$, show that $K \leq N_G(H)$.
Proof-
$N_G(H) = \{g\in G | gHg^{-1} = H\}$. K is nonempty being a subgroup of G. Since H is normal in K, and K is a subgroup of G, $\forall k \in K \Rightarrow k\in G$ we have $kHk^{-1} = H$ so that $k\in N_G(H)$. Thus K is a subgroup of $N_G(H)$.

3) If a group G contains an element a having exactly 2 conjugates, then G has a proper normal nontrivial subgroup N.
Proof-
Let G be a group and let $a\in G$ have exactly 2 conjugates. That is, there are 2 elements in the conjugacy class (orbit) $\bar{a} = \{gag^{-1} | g\in G\}$ of a. By theorem 4.3, $|\bar{a} | = 2 = | \{gag^{-1} | g\in G\}| = [G:G_a]$, but by previous theorem this gives that $G_a$ is normal in G. If $G_a$ were to be trivial then that would mean that G has 2 elements so is the same as $\mathbb{Z}_2$ and so is abelian. But in an abelian group, every conjugacy class has 1 element so this is a contradiction. Thus $G_a = N$ is nontrivial and normal in G.

4) Let H be a subgroup of G. Show that $C_G(H) \leq N_G(H)$
Proof-
We have that $N_G(H) = \{g\in G | gHg^{-1} = H\}$ and $C_G(H) = \{ g\in G |gh = hg \forall h \in H \}$. Note that the identity is in $C_G(H)$ so it is nonempty. Let $x\in C_G(H) \Rightarrow x\in G, hx = xh \forall h \in H$. We need to show that $xHx^{-1} = H$, which we do by subset inclusion. Let $y\in xHx^{-1}$ so that $y = xh_0x^{-1}$ for some $h_0 \in H$. Since $hx = xh \forall h \in H$ we have that $h_0x = xh_0 \Rightarrow xh_0x^{-1} = h_0 = y \in H$. Now for the other inclusion let $y \in H$. Then since $x\in C_G(H)$ we have $yx = xy \Rightarrow y =xyx^{-1} \in xHx^{-1}$ and thus $xHx^{-1} = H$.
This gives us that if $x\in C_G(H) \Rightarrow x \in N_G(H)$ so that $C_G(H) \leq N_G(H)$.
In fact we can also say that $C_G(H) \triangleleft N_G(H)$. Let $c\in C_G(H)$ and $n\in N_G(H)$. We must show that $cnc^{-1} \in C_G(H)$. Let $h\in H$ and note we have $nHn^{-1} = H$ so that $h = nh'n^{-1}$ for some $h' \in H$ and then $h' = n^{-1} hn$. Finally we have:
$(ncn^{-1})h = ncn^{-1}nh'n^{-1} = nch'n^{-1} = nh'cn^{-1} = nn^{-1}hncn^{-1} = h(ncn^{-1})$ so that $C_G(H) \triangleleft N_G(H)$.

5) If H is a subgroup of G, the factor group $N_G(H) / C_G(H)$ is isomorphic to a subgroup of $AutH$.
Proof-
First we define a map, $N_G(H) / C_G(H) \times H \rightarrow H$ by $(nC_G(H),h) \mapsto nhn^{-1}$. This map is a group action so it induces a group homomorphism $\phi : N_G(H)/C_G(H) \rightarrow AutH$ given by $nC_G(H) \mapsto \tau_{nC_G(H)}$, where $\tau_{nC_G(H)} : H \rightarrow H$ is given by $\tau_{nC_G(H)}(h) = nhn^{-1}$. Next we need to find $Ker\phi$. For $x\in N_G(H)$, we have that:
$xC_G(H) \mapsto 1_H \text{ iff } \tau_{xC_G(H)}(h) = h \forall h\in H$ $\text{ iff } xhx^{-1} = h \forall h \in H \text{ iff } xh = hx \forall h \in H$ $\text{ iff } x \in C_G(H) \text{ iff } xC_G(H) = C_G(H)$. Thus $Ker\phi = C_G(H)$ so that $\phi$ is one to one. It is onto $Im\phi$ which is automatically a subgroup. Thus $N_G(H) / C_G(H)$ is isomorphic to a subgroup of $AutH$.

6) Let G be a group acting on a set S containing at least 2 elements. Assume that G is transitive, i.e. that given any $x,y \in S$ there exists $g\in G$ such that $gx = y$. Prove:
a) For any $x\in S$, the orbit of x is in S.
Proof-
Let $x\in S$. Then the orbit of x is the set $\{ gx|x\in G\}$. Since $G\times S \rightarrow S$ is a group action we have that any element g acting on x is in S. Thus $\{ gx|x\in G\} \subset S$. Let $y \in S$. Then by transitivity of G there is some $g\in G$ such that $gx = y$, so that $y \in \{ gx|x\in G\}$. Thus $S \subset \{ gx|x\in G\}$ so in fact $S = \{ gx|x\in G\}$.

b) All the stabilizers $G_x$ are conjugate.
Proof-
Let $G_x,G_y$ be two arbitrary stabilizers, we will show they are conjugate. It is given that there exists $g\in G$ such that $gx=y$. Thus $g^{-1}y= x$. Claim that $G_x = g^{-1}G_yg$ by subset inclusion. Let $z\in G_x$. Then $zx=x$. Note that $z = g^{-1}(gzg^{-1})g$ and since $gzg^{-1}y = gzx = gx = y$ we know that $gzg^{-1} \in G_y$. Thus $z\in g^{-1}G_yg$. Next let $z\in g^{-1}G_yg$ so that we have $z = g^{-1}wg$ for some $w\in G_y$. Then $zx = g^{-1}wgx = g^{-1}wy = g^{-1}y = x$ tells us that $x\in G_x$. Thus they are conjugate.

7) Let G be a group and let Inn(G) be the set of all inner automorphisms of G. Show that $InnG \triangleleft Aut G$.
Proof-
From the definition of the inner automorphisms, we automatically get that $InnG \leq AutG$. Now to show it is normal, let $\phi \in AutG$ and let $\tau_g \in InnG$ be the inner automorphism induced by an arbitrary $g\in G$. For a given $x\in G$ we have:
$\phi \tau_g \phi^{-1} (x) = \phi \tau_g (\phi^{-1} (x)) = \phi ( g \phi^{-1} (x) g^{-1}) = \phi (g) \phi ( \phi^{-1} (x)) \phi(g^{-1} ) = \phi (g) x \phi (g)^{-1}$. This element is the inner automorphism induced by $\phi (g)$. I.e. $\tau_{\phi (g)} : G\rightarrow G$where $\tau_{\phi (g)}(x) = \phi (g) x \phi (g)^{-1}$. Thus we have shown normality.

9) Prove that if G is a group and $G/C(G)$ is cyclic then G is abelian.
Proof-
Suppose that $G/C(G)$ is cyclic. It is a set of cosets so to be cyclic means to be generated by a coset, i.e, $G/C(G) = \langle gC(G) \rangle$ for some element g in G. Then for two elements $g_1,g_2 \in G$ we have that:
$g_1C(G) = g^rC(G)$
$g_2C(G) = g^sC(G)$
This gives that $g_1g^{-r} = x \in C(G)$ and $g_2g^{-s} = y \in C(G)$. Then finally $g_1g_2 = xg^ryg^s = xyg^rg^s = yg^sxg^r = g_2g_1$ so that G is abelian.

11) Let G be a group containing an element a not of order 1 or 2. Show that G has a nonidentity automorphism.
Proof-
First, suppose that G is abelian. Then we have that $x\mapsto x^{-1}$ is an automorphism of G. The only way this automorphism can be the identity map is if G is trivial or if $x^{-1} = x \forall x \in G$. Since $a\in G$ we know that G is nontrivial. The second conclusion gives that $x^2 = e \forall x\in G$, but we know a does not have order 1 or 2, so this is false. Thus the automorphism is not the identity map.
If G is not abelian then we have $x \mapsto gxg^{-1}$ for each $g\in G$ is an automorphism of G. If it is the identity then we have that $gxg^{-1} = x \forall g,x\in G \Rightarrow gx=xg \forall g,x\in G$. This tells us that G is abelian, which is a contradiction to the assumption. This the map is a nonidentity automorphism.