Algebra Question from Mathematics GRE

Question

I just started learning algebra, and I came across a question from a practice GRE which I couldn't solve. http://www.wmich.edu/mathclub/files/GR8767.pdf #49

The finite group $G$ has a subgroup $H$ of order 7 and no element of $G$ other than the identity is its own inverse. What could the order of $G$ be?
Edit: This is a misreading of the problem. The problem intends that no element in G is its own inverse.

a) 27
b) 28
c) 35
d) 37
e) 42

I've already eliminated a) and d) due to LaGrange's theorem.

Answer 1

The order of a subgroup always divides the order of a group. You can immediately rule out a) and d). We're not allowed to have elements of order 2, but we would have them if the group was of even order. Therefore c).

Answer 2

Since $\rm a=a^{-1}\iff a^2=1$, you can use Cauchy's theorem to eliminate even orders.

Answer 3

Hint: You want to make sure there is no element of order $2$.

Answer 4

Because no element is its own inverse (other than the identity), then for every element $a$ (other than the one element which is the identity) there must exist an element $b$ such that $a^{-1}=b$. You could almost think of these elements as coming in pairs. If we ignore the identity, then $|G|-1$ has to be even...which eliminates every other choice except (c).