This question was posed in a class recently. We're not very "far" into group theory so I don't think it requires much specialized knowledge...

H = { A | A^2 = I, A in GL(2,7) }

(GL(2,7) is the general linear group of invertible 2x2 matrices with entries from the field Z_7; H is the subgroup of all elements of GL(2,7) with order 2)

What is the order of H?

I took an element A of H,

A=[a, b; c, d]

and squared it, set it to I to get

a^2 + bc = 1

ab + bd = 0

ac + cd = 0

bc + d^2 = 1

This seems a likely spot for me to have made a mistake. I split this system into 4 cases of solutions.

1) b = 0, c = 0, a^2 = 1, d^2 = 1

This case includes 4 solutions, with a and d chosen from {1,6} (since 6 is "negative 1")

2&3) b XOR c is zero. a^2=1 and d=-a

24 solutions (choose b XOR c, choose a value for it, choose a from {1,6}: 2x6x2), distinct from above where b=c=0.

4) b,c both nonzero.

d=-a again, but now

a^2 + bc = 1, rewritten: c=(1-a^2)/b. Since c is nonzero, we get the restriction that 1-a^2 != 0, so a is neither 1 nor 6.

30 more distinct solutions, from the 5 choices of a and 6 choices of b.

So my initial answer is 58 elements. Unfortunately for me, we were given a "hint" from a putnam exam that basically says the order of H is odd unless I misunderstand:

Suppose a finite group (like GL(2,7)) has exactly n elements of order p (I claim 58 elements of order 2). Then either n=0 or p divides n+1.

Somewhere I missed something. Any thoughts?