My Math Forum Isomorphic Groups..U(8) and Z4

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 March 30th, 2011, 05:06 PM #1 Newbie   Joined: Mar 2011 Posts: 6 Thanks: 0 Isomorphic Groups..U(8) and Z4 Prove or disprove U( isomorphic to Z4 U( = {1,3,5,7) Z4= {0,1,2,3}
 March 30th, 2011, 06:04 PM #2 Senior Member   Joined: Aug 2010 Posts: 195 Thanks: 5 Re: Isomorphic Groups..U(8) and Z4 How many elements of order 2 are in $U_8$?
 March 31st, 2011, 05:40 AM #3 Newbie   Joined: Mar 2011 Posts: 6 Thanks: 0 Re: Isomorphic Groups..U(8) and Z4 3 I believe. Does that make it non-isomorphic?
March 31st, 2011, 07:23 AM   #4
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Re: Isomorphic Groups..U(8) and Z4

Quote:
 Originally Posted by jcrot30 3 I believe. Does that make it non-isomorphic?
All four have order 2, since
1 = 1
9 = 1
25 = 1
49 = 1 (mod 8)

But this is not the case in Z4.

Have you considered Z2 x Z2 ??

 March 31st, 2011, 08:39 AM #5 Senior Member   Joined: Aug 2010 Posts: 195 Thanks: 5 Re: Isomorphic Groups..U(8) and Z4 As The Chaz points out, every element of $U_8$ has the property that $a^2= 1$ (the identity). Isomorphisms send the identity to the identity and respect the operation, so if $\alpha : U_8 \to \mathbb{Z}_4$ were an isomorphism, we would have: $\alpha(a*a)= \alpha(a) + \alpha(a)$ and $\alpha(1)= 0$ so $\alpha(a) + \alpha(a)= 0$ for all $a \in U_8$. But since $\alpha$ would be surjective, then every element $b \in \mathbb{Z}_4$ would satisfy $b+b= 0$. Is this true? (as a side note, technically you are correct jcrot30, that there are only 3 elements of order 2, since 1 is of order 1)
March 31st, 2011, 08:51 AM   #6
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Re: Isomorphic Groups..U(8) and Z4

Quote:
 Originally Posted by Turgul ... (as a side note, technically you are correct jcrot30, that there are only 3 elements of order 2, since 1 is of order 1)

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