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.
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).
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.
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)
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.
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.
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.
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.
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.
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.
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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