## 8767: Q49

Forum for the GRE subject test in mathematics.
YKM
Posts: 45
Joined: Mon Aug 08, 2011 10:25 am

### 8767: Q49

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?

I am a bit confused with the red part and how the answer was obtained, can anyone explain to me?

Thanks

blitzer6266
Posts: 61
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.

YKM
Posts: 45
Joined: Mon Aug 08, 2011 10:25 am

### Re: 8767: Q49

Thanks, Yes 35 is the only odd multiple of 7.