If the finite group G contains a subgroup of order 7 but no element (other than the identity) is its own inverse, then the order of G could be?
The answer is (C) 35.
I am a bit confused with the red part and how the answer was obtained, can anyone explain to me?
Thanks
8767: Q49

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 Joined: Sun Apr 04, 2010 1:08 pm
Re: 8767: Q49
If there is a subgroup of order k inside a group of order n, then k divides n.
Thus, in our group, 7 divides the order.
We can also deduce that the order is not even as follows: Suppose it were even. Then by Cauchy's theorem (if p prime divides n, then G has an element of order p), our group has an element of order 2. But an element of order 2 is a g such that g^2 = e. This contradicts our assumption (it is its own inverse).
I'm guessing 35 was the only of the choices that was odd and a multiple of 7.
Thus, in our group, 7 divides the order.
We can also deduce that the order is not even as follows: Suppose it were even. Then by Cauchy's theorem (if p prime divides n, then G has an element of order p), our group has an element of order 2. But an element of order 2 is a g such that g^2 = e. This contradicts our assumption (it is its own inverse).
I'm guessing 35 was the only of the choices that was odd and a multiple of 7.
Re: 8767: Q49
Thanks, Yes 35 is the only odd multiple of 7.