Throughout this post, and are the group of units and the Jacobson radical of a ring Assuming that is finite and is odd, we will show that for some positive integers Let’s start with a nice little problem.

**Problem 1**. Prove that if is finite, then is finite too and

**Solution**. Let and define the map by This map is clearly a well-defined group homomorphism. To prove that is surjective, suppose that Then for some and hence implying that So is surjective and thus

Now, is a subgroup of and Thus is finite and

**Problem 2**. Let be a prime number and suppose that is finite and Prove that if then

**Solution**. Suppose that and Then, considering as an additive group, is a subgroup of and so But then by Problem 1, and that’s a contradiction!

There is also a direct, and maybe easier, way to solve Problem 2: suppose that there exists On define the relation as follows: if and only if for some integer Then is an equivalence relation and the equivalence class of is Note that because and So if is the number of equivalence classes, then contradiction!

**Problem 3**. Prove that if is a finite field, then In particular, if is odd, then and is a power of

**Solution**. The group is isomorphic to the group of invertible linear maps Also, there is a one-to-one correspondence between the set of invertible linear maps and the set of (ordered) bases of So is equal to the number of bases of Now, to construct a basis for we choose any non-zero element There are different ways to choose Now, to choose we need to make sure that are not linearly dependent, i.e. So there are possible ways to choose Again, we need to choose somehow that are not linearly dependent, i.e. So there are possible ways to choose If we continue this process, we will get the formula given in the problem.

**Problem 4**. Suppose that is finite and is odd. Prove that for some positive integers

**Solution**. If in then would be a subgroup of order 2 in and this is not possible because is odd. So Hence and Let be the ring generated by and Obviously is finite, and We also have by Problem 2. So is a finite semisimple ring and hence for some positive integers and some finite fields by the Artin-Wedderburn theorem and Wedderburn’s little theorem. Therefore The result now follows from the second part of Problem 3.