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Old 02-06-2005, 11:11 PM   #1 (permalink)
vvaann
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Group order

I have no idea on how to solve this question. Please help me!

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 should be

(A) 27
(B) 28
(C) 35
(D) 37
(E) 42

Answer key:
SPOILER: (C)
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Old 02-07-2005, 02:53 PM   #2 (permalink)
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Re: Group order

Since G has a subgroup of order 7, 7 must divide |G|.
If you look at the prime factorization,
27=3x3x3, so no subgroup of order 7
28=2x2x7, so it COULD have an element that is its own inverse
35=5x7
37 is prime, so no subgroup of order 7
42=2x3x7, so it COULD have an element that is its own inverse

Someone please confirm rationale behind 28 and 42...
35 by process of elimination!
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Old 02-07-2005, 04:18 PM   #3 (permalink)
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Re: Group order

28 and 42 can be eliminated because 2 divides both numbers and therefore the corresponding group contains an element of order 2 (contradicting the hypothesis that there is no such element).
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Old 02-07-2005, 04:27 PM   #4 (permalink)
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Re: Group order

Great, mathie!
Now I'm clear about this question.

Since G has no element which is its own inverse, G has no subgroup of order 2 (the element and its inverse form a subgroup!). Hence, |G| is not dividable by 2.

In addition, as you indicated, |G| is dividable by 7.

Therefore |G| should be 35.
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Old 02-11-2005, 07:48 AM   #5 (permalink)
ish
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Re: Group order

the answer should be an odd number AND a multiple of 7, cos only the identity has a self inverse right?? so each other element would come in pairs, along with its inverse...so an odd multiple of 7, thats 35...
thats how i thot it would work out...
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Old 02-11-2005, 09:01 AM   #6 (permalink)
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Re: Group order

ish,

An element which has an inverse of itself is not neccessarily the identity.

For example, consider the group (under multiplication) of two elements: {1,-1}
Here, -1 has an inverse of itself, but it's not the identity.

Br,
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Old 02-11-2005, 04:52 PM   #7 (permalink)
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Re: Group order

I think what ish means is that in this problem, only the identity is its own inverse, so there must be an odd number of elements in the group. that is the way I did it too, but I am not really convinced by that reasoning, piero expresses it better.
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Old 02-11-2005, 06:32 PM   #8 (permalink)
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Re: Group order

guys can you plz tell me what actually the terms order and inverse mean.iam not able to figure it out,may be forgotten or known by another name or never known.i wud be really thankful.cya.bye.
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Old 02-11-2005, 07:03 PM   #9 (permalink)
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Re: Group order

??? Are you studying for the General GRE or the Math Subject GRE? If the former, don't worry about it, you don't need to know. If the latter, haven't you taken abstract algebra?

Do you know what a group is? The group axioms require that each element have an inverse, that is if G is a group, and a is an element of G, then there exists an element b in G such that ab = e = ba, where e is the group identity. The order of a group is the number of elements in the group.

C
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