Show that $GL_n(\mathcal{F})$ is a finite group if and only if $\mathcal{F}$ has a finite number of elements.

These are my thoughts.

The order of the group $GL_n(\mathcal{F})$ is $|GL_n(\mathcal{F})| = n^2$ because the matrices in the set are $n.n$ matrices and which would mean each matrix has $n^2$ entries. I thought of somehow arguing $\mathcal{F}$ has $n^2$ elements which would show it's finite. Perhaps I need to show $\mathcal{F}$ is a finite. Could you clarify if I'm on the right track or not.
Could you help get started on th right track on what the main idea of this proof? I'm stuck and don't if my assumptions are right and how to use them.

  • $\begingroup$ For an "if and only if" proof, you must break the proof into two parts. Here, for the first part, assume that F is a finite field and show that GL_n(F) has finite order. That is step one. Then for the second step assume GL_n(F) has finite order and show that F is finite. That is step two. That will complete the proof. $\endgroup$ – Tyler Levasseur Jun 26 '15 at 5:57

To prove the if and only if, we break the proof into two parts.

First assume the "right-hand side" $-$ that $\mathcal F$ has a finite number of elements (and let that number be represented by $|\mathcal F|$). Then what is the maximum possible number of elements for $GL_n(\mathcal F)$? That is, how do we count up all the different possible combinations? This gives an upper bound for the true number of elements, and is most certainly finite.

Now we'll assume the "left-hand" side to prove the "right-hand" side. Notice that $\mathrm{diag}_n(f_j) \in GL_n(\mathcal F)$ for any non-zero $f_j \in \mathcal F$, since the matrix is diagonal and $\mathcal F$ is a field (so every element has an inverse). Since $\{\mathrm{diag}_n(f_j) : f_j \in \mathcal F, f_j \neq 0\} \subset GL_n(\mathcal F)$ and $|GL_n(\mathcal F)| < \infty$, it follows that $|\mathcal F| < \infty$ also.

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    $\begingroup$ You need $f_i \neq 0$, but that's jsut a minor detail. $\endgroup$ – Arthur Jun 26 '15 at 6:23

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