My Math Forum Show that group Z2 x Z2 is not isomorphic to the group Z4

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 May 16th, 2012, 01:08 AM #1 Senior Member   Joined: Apr 2012 Posts: 106 Thanks: 0 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
 May 16th, 2012, 07:31 AM #2 Senior Member   Joined: Apr 2012 Posts: 106 Thanks: 0 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?
 May 16th, 2012, 01:58 PM #3 Senior Member   Joined: Jul 2011 Posts: 227 Thanks: 0 Re: Show that group Z2 x Z2 is not isomorphic to the group Z The order of $|\mathbb{Z}_2 \times \mathbb{Z}_2|=|\mathbb{Z}_2|\cdot |\mathbb{Z}_2|= 4$ 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 $f: \mathbb{Z}_2 \times \mathbb{Z}_2 \to \mathbb{Z}_4$ which is not bijective.
 May 16th, 2012, 02:37 PM #4 Senior Member   Joined: Apr 2012 Posts: 106 Thanks: 0 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.
 May 17th, 2012, 06:08 PM #5 Senior Member   Joined: Mar 2012 Posts: 294 Thanks: 88 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.
 May 28th, 2012, 11:16 AM #6 Newbie   Joined: May 2012 Posts: 3 Thanks: 0 Re: Show that group Z2 x Z2 is not isomorphic to the group Z $\mathbb{Z}_2 \times \mathbb{Z}_2$ dosn't have an element of order 4 ($|(0,0)|=1$, $|(0,1)|=|(1,0)|=|(1,1)|=2$). But $\mathbb{Z}_4$ does (for example, $|1|=4$). So, they are non-isomorphic. P.S.: $|x|$ - order of the element $x$.
 June 5th, 2012, 03:58 PM #7 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Show that group Z2 x Z2 is not isomorphic to the group Z To me the most obvious point is that $Z_2\times Z_2$ has the property that a+ a= 0 for all a. That is not true for $Z_4$.

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