GR8767 Q49 (Group Theory)

Forum for the GRE subject test in mathematics.
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speedychaos4
Posts: 47
Joined: Mon Mar 22, 2010 2:42 am

GR8767 Q49 (Group Theory)

Post by speedychaos4 » Sun Oct 17, 2010 3:55 am

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.

fireandgladstone
Posts: 27
Joined: Sun Oct 17, 2010 4:57 am

Re: GR8767 Q49 (Group Theory)

Post by fireandgladstone » Sun Oct 17, 2010 5:07 am

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.

le6tan
Posts: 8
Joined: Sat Apr 10, 2010 8:30 am

Re: GR8767 Q49 (Group Theory)

Post by le6tan » Sun Oct 17, 2010 5:45 am

We know that there is a subgroup of order 7. The fact that no element other than the identity-e 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.

speedychaos4
Posts: 47
Joined: Mon Mar 22, 2010 2:42 am

Re: GR8767 Q49 (Group Theory)

Post by speedychaos4 » Sun Oct 17, 2010 7:07 am

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...



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