Proving U(8) is not Isomorphic to U(10): Insights and Techniques

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SUMMARY

The discussion focuses on proving that the group U(8) is not isomorphic to the group U(10). Key observations include that every element in U(8) is its own inverse, while in U(10), the elements 3 and 7 are inverses of each other. The proof can be formalized by demonstrating that an isomorphism would require mapping inverses in a way that leads to a contradiction, specifically that f(7) must equal f(3), which is impossible. This conclusion is supported by the properties of group homomorphisms and the structure of the groups involved.

PREREQUISITES
  • Understanding of group theory concepts, specifically the structure of unit groups U(n).
  • Familiarity with the properties of isomorphisms in algebraic structures.
  • Knowledge of Cayley tables and their use in group theory.
  • Experience with automorphisms and their implications in group mappings.
NEXT STEPS
  • Study the properties of unit groups, specifically U(n) for various values of n.
  • Learn about group homomorphisms and their role in establishing isomorphisms.
  • Explore the concept of inverses in groups and how they affect group structure.
  • Investigate the use of Cayley tables for proving group properties and relationships.
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Mathematicians, students of abstract algebra, and anyone interested in group theory and its applications in proving isomorphism between algebraic structures.

k3N70n
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Hi. Hoping a could have a little bit of guidance with this question

Show that U(8) is not isomorphic to U(10)

So far, I've realized that in U(8) each element is it's own inverse while in U(10) 3 and 7 are inverses of each other. I guess that's really all I need to say that they aren't isomorphic but my suspicion is that I should be stating this in a more formal way then a simple Cayley table.

Previously, I worked out a somewhat similar question where I had to find the automorphisms of Z_4. I said:
let f:Z_4 --> Z_4
and f(0)=0 (because the identity must be mapped to the identity by a theorem early proved)
then f(2) = f(1) + f(1)
f(3) = f(1) + f(2)

So then we have 4 cases for f(1)...[went on to show that if f(1) = 0 or 2 then f was not injective]

I was thinking something similar here may be appropriate but I'm not sure how to set it up. Thanks in advance for any help
 
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I see nothing wrong with saying exactly what you did. If there exist an isomorphism from U(10) to U(8), then f(7) and f(3) must map into inverses: but what ever f(10) is, its inverse is itself: f(7) must equal f(3) contradicting the fact that it is an isomorphism.
 
Thanks HallsofIvy. I guess I was just trying to make the question harder than it really was.
 

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