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# Abstract Algebra: Direct Products

Posted on 2008-10-29
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I have two problems that i need help with! I'm trying to study for an exam and cannot figure out these review problems

first problem:
The group S3 (+) Z2 is isomorphic to one of the following groups: Z12,
Z6 (+) Z2, A4, D6. Determine which one by elimination.

I know to be isomorphic denoted phi from a group G to a group G' is a one to one mapping from G onto G' that preserves the group operation. So phi(ab)= phi(a) phi(b). I also know to prove one is isomorphic, need to have mapping (define a function phi from G to G'), need to be 1 to 1 (assume that phi(a)=phi(b) and prove that a=b), needs to be onto (prove for any element g' in G', find an element g in G such that phi(g)=g'), and prove that phi is operation preserving (show phi(ab)= phi(a)phi(b)).

I also know that S3 (+) Z2 the first group S3 is multiplicative and second group Z2 is additive so dealing with mixed binary operations.
I think S3 (+) Z2 = {(1), 0), ((1 2), 0), ((1 2 3), 1)}
then ((1 2, 0) * ((1 2 3), 1)= ((1 2)(1 2 3), 1)= ((1)(2 3 1), 1). But i'm not too sure.

second problem:
Find a subgroup of Z12 (+) Z4 (+) Z15 that is of order 9.

I know that the direct product is the set of all n-tuples for which the ith componenet is an element of Gi and the operation is componentwise.

I'm confused how to find direct products, can someone help me? Thanks so much, greatly appreciated!
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Question by:dongyowlin
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LVL 84

Accepted Solution

ozo earned 500 total points
A4

Z3 (+) Z1 (+) Z3
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Author Comment

how does A4 equal to Z3 (+) Z1 (+) Z3?
A i know means alternating group. but im still confused...

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LVL 84

Expert Comment

A4 is isomorphic to  S3 (+) Z2

Z3 (+) Z1 (+) Z3 is a subgroup of Z12 (+) Z4 (+) Z15 that is of order 9.
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LVL 20

Expert Comment

Unfortunately, all candidates have the correct order (12), so this cannot be used for elimination.
Since S3 is not abelian, the direct product isn't abelian either.
This eliminates Z12 and Z2(+)Z6.
The only subgroup of order 3 in D6 consists of the rotations with multiples of 120°.
Such a rotation never commutes with any element of order 2 (i.e. reflection); this eliminates D6.

2)
Z12 and Z15 have a subgroup isomorphic to Z3 each, together they make a group of order 9.
I.e., the subgroup generated by (4,0,0) and (0,0,5) is what you are looking for.
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