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I have seen (in Dummit and Foote - Section $1.6$) that the cardinality of the group $GL_{2}(\mathbb{F}_2)$ equals cardinality of the group $S_3$ (since they are isomorphic) and I'm trying to verify it for the former. I'm just having trouble counting it.

I get that $GL_{2}(\mathbb{F}_2)$ is the group of matrices where $ad-bc \neq 0$ and where the elements of each matrix comes from the finite field $\mathbb{F}_2$ of just $2$ elements.

The total number of $2 \times 2$ matrices possible with elements from $\mathbb{F}_2$ is $16$ (two choices per matrix entry).

I want to count the number of ways $ad-bc\neq 0$ and minus it from $16$ to get $6$ and all my counts are coming up short of $10$. Any assistance would be great.

Samuel Adrian Antz
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Salazar_3854708
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4 Answers4

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Elements of $GL_2(\mathbb{F}_2)$ correspond to a choice of 2 independent vectors in $(\mathbb{F}_2)^2$. There are $2^2 - 1$ choices for the first vector $v$, since it cannot be in $span() = 0$. Once we’ve chosen $v$, there are $2^2 - 2$ choices for the second vector, since it cannot be in $span(v_1)$, a space with 2 elements. So there are a total of $(2^2 - 1)(2^2 - 2) = 6$ elements.

Mark Saving
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For $ad-bc\ne0$ in $\Bbb F_2$, one of the products $ab$ must be $0$ and the other must be $1$. There are $2$ ways to pick which product should be $1$. Product one is only possible in one way: $1\cdot 1$. The other $3$ possible products are $0$. That’s $2\cdot 1\cdot 3=6$.

Hagen von Eitzen
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$1111,0101,0100,0110,1010,1000,1001,0001,0010,0000 $are the ten ways the string $adbc$ can be for the equation $ad=bc$ to hold.

The other $6$ strings give an element of the general linear group each.

Umesh Shankar
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To give a more general answer to this problem as there are already three answers explaining how to solve this particular problem and you could find this useful:

The cardinality of $\operatorname{GL}(m,\mathbb Z_n)$ can be given as a formula dependent on $m$ and $n$ using the Jordan function: $$J_k(n) :=n^k\prod_{p\in\mathbb{P},p\mid n}\left(1-\frac{1}{p^k}\right)$$ and is given by: $$|\operatorname{GL}(m,\mathbb Z_n)| =n^{\frac{(m-1)m}{2}}\prod_{k=1}^mJ_k(n).$$ (For $\operatorname{SL}(m,\mathbb Z_n)$, the product starts at $k=2$, so the product with the Euler function $J_1(n)=\varphi(n)=|\mathbb Z_n^\times|$ is missing. This corresponds to the fact, that $\operatorname{GL}(m,\mathbb Z_n)/\operatorname{SL}(m,\mathbb Z_n)\cong \mathbb Z_n^\times$.)

Samuel Adrian Antz
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