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I would like to know if my proof below is correct.

Problem Show that $GL_n(\mathbb{F})$ is a finite group iff $\mathbb{F}$ has a finite number of elements.

Solution

  • If $\mathbb{F}$ is a finite field and say $\vert \mathbb{F} \vert = m$, then the set of all $n \times n$ matrices, which is a superset of $GL_n(\mathbb{F})$, where the elements comes from $\mathbb{F}$ has cardinality $m^{n^2}$. Hence, $\vert GL_n(\mathbb{F}) \vert < m^{n^2}$ and thereby $GL_n(\mathbb{F})$ is a finite group.

  • If $GL_n(\mathbb{F})$ is a finite group, note that for every non-zero element $a \in \mathbb{F} -\{0\}$, the matrix $aI_{n \times n}\in GL_n(\mathbb{F})$, where $I_{n \times n}$ is the identity matrix with the multiplicative identity along the diagonal. This means $\vert \mathbb{F} \vert \leq GL_n(\mathbb{F}) + 1$ and thereby $\mathbb{F}$ is a finite field.

Thanks

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    $\begingroup$ Looks good to me. $\endgroup$ – Gerry Myerson Dec 27 '13 at 19:02
  • $\begingroup$ Thanks Jerry. That was really quick! $\endgroup$ – John Smith Dec 27 '13 at 19:03
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    $\begingroup$ "and thereby $GL_n(\mathbb F)$ is a finite field..." I think you mean to say "finite group" here. $\endgroup$ – Calle Dec 27 '13 at 19:04
  • $\begingroup$ Yes, Calle you are correct. I have changed it now. $\endgroup$ – John Smith Dec 27 '13 at 19:06
  • $\begingroup$ It's an organizational issue rather than a factual one, but you could just show that there are inclusions of sets $\mathbb{F}^{\times} \to GL_n(\mathbb{F}) \to \mathbb{F}^{n^2}$ (which you've written out above). $\endgroup$ – anomaly Mar 8 '16 at 16:43
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Looks great. The only adjustment I'd make is to say that $\left|GL_n(\Bbb F)\right|\le m^{n^2}.$ It's certainly true that $\left|GL_n(\Bbb F)\right|<m^{n^2},$ but you haven't justified it, and the strictness isn't necessary for finiteness. That's largely an aesthetic issue, though, not a technical one.

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  • $\begingroup$ Yes, you are right. I need to add that $0 \cdot I_n$ does not belong to $GL_n(\mathbb{F})$ and now the strictness should hold. But either way it is not required. Thanks. $\endgroup$ – John Smith Dec 27 '13 at 19:44
  • $\begingroup$ Indeed, that's a perfect justification. You're welcome. $\endgroup$ – Cameron Buie Dec 27 '13 at 19:46
  • $\begingroup$ Thanks. At last a well behaved, polite user on this site. $\endgroup$ – John Smith Dec 27 '13 at 19:51
  • $\begingroup$ I can but try. ;-) Sorry to hear that your reception hasn't been better so far. $\endgroup$ – Cameron Buie Dec 27 '13 at 19:52

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