Algebra: Non-isomorphic groups

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Homework Help Overview

The discussion revolves around the question of how many non-isomorphic groups of two elements exist, particularly in the context of group theory. Participants explore the properties of groups with two elements and the conditions for isomorphism.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants attempt to understand the definition of non-isomorphic groups and explore the construction of groups with two elements under different operations. Questions arise about the possibility of generating non-isomorphic groups and the general method for determining the number of non-isomorphic groups for any number of elements.

Discussion Status

The discussion includes attempts to clarify the original question and the properties of groups of order two. Some participants express uncertainty about the feasibility of creating non-isomorphic groups, while others suggest that all groups of order two may be isomorphic to the cyclic group of order two.

Contextual Notes

Participants note the challenge of defining non-isomorphic groups and the implications of group homomorphisms in this context. There is an acknowledgment of the complexity involved in determining non-isomorphic groups for larger sets.

nuuskur
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Homework Statement


How many non-isomorphic groups of two elements are there?

Homework Equations

The Attempt at a Solution


I don't understand exactly what we are being asked.
If we have a group of two elements under, say, addition, then G =\{0, g\}.
Then also g+g = 0 must be true, means g is its own opposite. (of order 2).

Now, how should I construct a group under some operation ##*##: G' = \{e, g'\}, where ##e## is the unit/zero element (depending on operation) such that G and G' are not isomorphic?

G\cong G' iff there exists a bijective group homomorphism f: G\to G'
I can define f such that:
f(0) = e (satisfies one of the group homormorphism requirement)
f(g) = g'
Is bijection and

f(0+g) = f(g) = g' = e*g'
f(g+g) = f(0) = e = g'*g'
so G\cong G'

What must I do to generate non-isomorphic groups of two elements?

In general, for any ##n##, how can I determine the number of non-isomorphic groups of ##n## elements?
 
Last edited:
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nuuskur said:
What must I do to generate non-isomorphic groups of two elements?
Do you think that is possible?

nuuskur said:
In general, for any ##n##, how can I determine the number of non-isomorphic groups of ##n## elements?
I think that is not easy at all.
 
nuuskur said:

Homework Statement


How many non-isomorphic groups of two elements are there?

Homework Equations

The Attempt at a Solution


I don't understand exactly what we are being asked.
If we have a group of two elements under, say, addition, then G =\{0, g\}.
Then also g+g = 0 must be true, means g is its own opposite. (of order 2).

Now, how should I construct a group under some operation ##*##: G' = \{e, g'\}, where ##e## is the unit/zero element (depending on operation) such that G and G' are not isomorphic?

G\cong G' iff there exists a bijective group homomorphism f: G\to G'
I can define f such that:
f(0) = e (satisfies one of the group homormorphism requirement)
f(g) = g'
Is bijection and

f(0+g) = f(g) = g' = e*g'
f(g+g) = f(0) = e = g'*g'
so G\cong G'

What must I do to generate non-isomorphic groups of two elements?

You can't. You are asked "how many non-isomorphic groups of two elements are there?" and you've shown that the answer is "all groups of order two are ismorphic to the cyclic group of order two."
 
EDIT: nope
 

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