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.

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One Response to Algebra Solutions: Hungerford II.4

  1. Alaa Yousof says:

    I need this question ( 15 hangurford )
    if a normal subgroup N of order p , ( p prime ) is contained in a group G of order p^n, then N is in the center of G

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