**1) Let G be a group and A a normal abelian subgroup. Show that operates on A by conjugation and obtain a homomorphism
Proof-** Define by and show it is a well defined map. Let . Then and . This gives that so that for some element in A. Then and . Then finally we have:

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 and that:

thus the map is a group action. This action induces a group homomorphism defined by where is given by

**2) If H, K are subgroups of G such that , show that .
Proof-** . K is nonempty being a subgroup of G. Since H is normal in K, and K is a subgroup of G, we have so that . Thus K is a subgroup of .

**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 have exactly 2 conjugates. That is, there are 2 elements in the conjugacy class (orbit) of a. By theorem 4.3, , but by previous theorem this gives that is normal in G. If were to be trivial then that would mean that G has 2 elements so is the same as and so is abelian. But in an abelian group, every conjugacy class has 1 element so this is a contradiction. Thus is nontrivial and normal in G.

**4) Let H be a subgroup of G. Show that
Proof-** We have that and . Note that the identity is in so it is nonempty. Let . We need to show that , which we do by subset inclusion. Let so that for some . Since we have that . Now for the other inclusion let . Then since we have and thus .

This gives us that if so that .

In fact we can also say that . Let and . We must show that . Let and note we have so that for some and then . Finally we have:

so that .

**5) If H is a subgroup of G, the factor group is isomorphic to a subgroup of .
Proof-** First we define a map, by . This map is a group action so it induces a group homomorphism given by , where is given by . Next we need to find . For , we have that:

. Thus so that is one to one. It is onto which is automatically a subgroup. Thus is isomorphic to a subgroup of .

**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 there exists such that . Prove:
a) For any , the orbit of x is in S.
Proof-** Let . Then the orbit of x is the set . Since is a group action we have that any element g acting on x is in S. Thus . Let . Then by transitivity of G there is some such that , so that . Thus so in fact .

**b) All the stabilizers are conjugate.
Proof-** Let be two arbitrary stabilizers, we will show they are conjugate. It is given that there exists such that . Thus . Claim that by subset inclusion. Let . Then . Note that and since we know that . Thus . Next let so that we have for some . Then tells us that . 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 .
Proof-** From the definition of the inner automorphisms, we automatically get that . Now to show it is normal, let and let be the inner automorphism induced by an arbitrary . For a given we have:

. This element is the inner automorphism induced by . I.e. where . Thus we have shown normality.

** 9) Prove that if G is a group and is cyclic then G is abelian.
Proof-** Suppose that is cyclic. It is a set of cosets so to be cyclic means to be generated by a coset, i.e, for some element g in G. Then for two elements we have that:

This gives that and . Then finally 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 is an automorphism of G. The only way this automorphism can be the identity map is if G is trivial or if . Since we know that G is nontrivial. The second conclusion gives that , 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 for each is an automorphism of G. If it is the identity then we have that . This tells us that G is abelian, which is a contradiction to the assumption. This the map is a nonidentity automorphism.

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