49. 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
A.27 B.28 C.35 D.37 E.42
I'm new to group theory so I'm not sure what is the proper approach to this question. I know from Lagrange's Th that 7 divides the order of G. And since no element in the subgroup is its own inverse, I guess it indicates that the subgroup is a cyclic subgroup. And then I'm stuck in the process.
Would anyone be willing to offer a tip or help me correct my approach? Thanks.
GR8767 Q49 (Group Theory)

 Posts: 27
 Joined: Sun Oct 17, 2010 4:57 am
Re: GR8767 Q49 (Group Theory)
The answer is (C). Take the additive group (Z/35Z, +) and consider the subgroup generated by 5. Clearly no element of (Z/35, +) other than the identity is its own inverse. Lagrange's theorem eliminates (A) and (D). Cauchy's theorem eliminates (B) and (E) since 2  28 and 2  42. Cauchy's theorem is generalized in one of the Sylow theorems.
Re: GR8767 Q49 (Group Theory)
We know that there is a subgroup of order 7. The fact that no element other than the identitye is it's own inverse means, that there are now element for which a=a^{1} <=> a^{2}=e <=> there is now subgroup of order 2. The only group order, which is coprime to 2 and also divides 7(Lagrange's theorem) is 35, or answer C.

 Posts: 47
 Joined: Mon Mar 22, 2010 2:42 am
Re: GR8767 Q49 (Group Theory)
Thanks fireandgladstone and le6tan...I made a little mistake in reading the question...I thought it mean no element in the subgroup has itself as an inverse(of course there isn't)...forgive my poor English...