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May 16th, 2012, 01:08 AM   #1
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Show that group Z2 x Z2 is not isomorphic to the group Z4

Hello dear colleagues,

I need to show that the group Z2 X Z2 is not isomorphic to group Z4. (Z is a whole number).

I would appreciate any offered help,
V
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May 16th, 2012, 07:31 AM   #2
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Re: Show that group Z2 x Z2 is not isomorphic to the group Z

I am completely new at this field - so I don't even get how do we get order from the group Z2 x Z2.
For example, whats the order of (1,0)? How do we calculate that?
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May 16th, 2012, 01:58 PM   #3
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Re: Show that group Z2 x Z2 is not isomorphic to the group Z

The order of
The order of a group is the cardinality of the group. Do you know the definition of the order of an element of a group?

To solve the exercice I think you need to find a function which is not bijective.
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May 16th, 2012, 02:37 PM   #4
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Re: Show that group Z2 x Z2 is not isomorphic to the group Z

I used the rule that order of elements in Z2 x Z2 must be equal to the order of elements in Z4. It worked.
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May 17th, 2012, 06:08 PM   #5
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Re: Show that group Z2 x Z2 is not isomorphic to the group Z

the standard proof goes something like this:

(0,1) in Z2 x Z2 has order 2:

(0,1) + (0,1) = (0,0) (since 1+1 = 2 = 0 (mod 2)).

(1,0) in Z2 x Z2 also has order 2:

(1,0) + (1,0) = (0,0).

now if Z2 x Z2 and Z4 were isomorphic, Z4 would contain (at least) 2 elements of order 2 (the images of (1,0) and (0,1) under the isomorphism).

but:

0 is of order 1

1+1 = 2
1+1+1 = 3
1+1+1+1 = 0 <---1 is of order 4

2+2 = 0 <---2 is of order 2

3+3 = 2 (since 6 = 2 (mod 4))
3+3+3 = 1 (since 9 = 1 (mod 4))
3+3+3+3 = 0 <---3 is of order 4.

since Z4 has only one element of order 2 (namely, 2), it cannot be isomorphic to Z2 x Z2.
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May 28th, 2012, 11:16 AM   #6
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Re: Show that group Z2 x Z2 is not isomorphic to the group Z

dosn't have an element of order 4 (, ). But does (for example, ). So, they are non-isomorphic. P.S.: - order of the element .
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June 5th, 2012, 03:58 PM   #7
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Re: Show that group Z2 x Z2 is not isomorphic to the group Z

To me the most obvious point is that [itex]Z_2\times Z_2[/itex] has the property that a+ a= 0 for all a. That is not true for [itex]Z_4[/itex].
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