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#1 2009-03-22 11:06:40

LuisRodg
Real Member
Registered: 2007-10-23
Posts: 322

Couple questions regarding isomorphisms

Some help (hints) greatly appreciated!

Last edited by LuisRodg (2009-03-22 11:11:40)

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#2 2009-03-22 12:14:52

Ricky
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Registered: 2005-12-04
Posts: 3,791

Re: Couple questions regarding isomorphisms

For the first, you want to be looking at generators of the group.  If you define where generators are sent, you define the entire isomorphism.

For the second, write x = g^k * z_1 and y = g^l * z_2 where z_1 and z_2 are in Z(G).  Now compute xy and yx.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#3 2009-03-22 22:56:06

JaneFairfax
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Registered: 2007-02-23
Posts: 6,868

Re: Couple questions regarding isomorphisms

(ii) has interesting consequences.


dizzy

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#4 2009-03-25 02:05:52

LuisRodg
Real Member
Registered: 2007-10-23
Posts: 322

Re: Couple questions regarding isomorphisms

Hello Ricky. Your hint for (ii) easily solves the problem as one can easily show that xy = yx, hence Abelian. However, I was wondering why you can write x and y in such form. Here is what I tried but I have just one question:

And here is where Im stuck. How can I deduce from the last line that:

If I can do so then the problem is done. Could you explain how to deduce such?

Thanks.

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#5 2009-03-25 02:30:15

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

Re: Couple questions regarding isomorphisms

No, no. Use the fact that the set of all cosets of a group splits the group into partitions – every element of a group belongs to exactly one coset. Hence

And since

Last edited by JaneFairfax (2009-03-25 03:04:14)

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#6 2009-03-25 02:39:19

LuisRodg
Real Member
Registered: 2007-10-23
Posts: 322

Re: Couple questions regarding isomorphisms

But I want to prove that G is Abelian, so dont I have to pick the arbitrary elements x,y in G and not in G/Z(G) ?

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#7 2009-03-25 03:06:06

JaneFairfax
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Registered: 2007-02-23
Posts: 6,868

Re: Couple questions regarding isomorphisms

Argh! I made a slight mistake. Edited and corrected now.

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#8 2009-03-25 10:15:34

LuisRodg
Real Member
Registered: 2007-10-23
Posts: 322

Re: Couple questions regarding isomorphisms

Thanks.

I still do not understand how to do (i).

"If you define where generators are sent, you define the entire isomorphism."

Could explain what you mean by that? and could you show me how to solve one of the problems in (i) so I have an idea of what you are saying as I feel completely lost.

Last edited by LuisRodg (2009-03-25 10:16:16)

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#9 2009-03-25 14:38:22

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: Couple questions regarding isomorphisms

I shall find the automorphism group of G = Z2 x Z4 = {(0, 0), (1, 0), (0, 1), (1, 1), (0, 2), (1, 2), (0, 3), (1, 3)}.  Addition is preformed component wise.  i.e. (1, 2) + (1, 3) = (1 + 1, 2 + 3) = (2, 5) = (0, 1).

This group is generated by (1, 0) and (0, 1).  To see this, (x, y) = (x, 0) + (0, y).  Now add (1, 0) to itself x times, and (0, 1) to itself y times.

Thus, if I have an isomorphism from G to itself, then as soon as I define where (1, 0) and (0, 1) go, I have defined the isomorphism.  This is because isomorphisms preserve addition.

So now I just need to figure out where I may send (1, 0) and (0, 1).  First off, isomorphisms preserve the order of elements.  (1, 0) has order 2, so it must be sent to an element of order 2.  Similarly with (0, 1) having order 4.

Elements of order 2: (1, 0), (1, 2)

Elements of order 4: (0, 1), (1, 1), (0, 3), (1, 3)

From this, there are a total of 8 possible maps.  All that remains is to check each one to verify it is an isomorphism.  While a bit tedious, there is nothing difficult going on here.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#10 2009-12-13 19:02:49

atahat
Guest

Re: Couple questions regarding isomorphisms

also (0,2) has order 2 in Z2XZ4

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