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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.

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The precise wording of the problem is "If the finite group $G$ contains a subgroup of order seven but no element (other than the identity) is its own inverse, then the order of $G$ could be". I don't think you are interpreting the problem correctly: The intention is that no element in $G$ is its own inverse. –  Alex Zorn Mar 12 '13 at 5:29
    
...No wonder I was having so much trouble. –  Mark Mar 12 '13 at 5:31
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4 Answers

up vote 5 down vote accepted

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).

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Now I see your LaGrange theorem comment, whoops. –  meh Mar 12 '13 at 5:10
    
Dear @meh, your LaGrange is related somehow to Lagrange? –  user26857 Aug 13 '13 at 10:55
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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).

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No the condition that no element is it's own inverse only applies to the subgroup of size 7. But we already knew that it's size was odd. –  Mark Mar 12 '13 at 5:22
    
When I clicked the link and read question 49, it was worded differently than proposed in the above. I think the question means to say that no element in $G$ is its own inverse. –  Brandon Mar 12 '13 at 5:33
    
Yes, sorry I was misinterpreting the question. –  Mark Mar 12 '13 at 5:33
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Since $\rm a=a^{-1}\iff a^2=1$, you can use Cauchy's theorem to eliminate even orders.

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

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