Algebra: Non-isomorphic groups

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There are no non-isomorphic groups of two elements; all groups of order two are isomorphic to the cyclic group of order two. The discussion highlights the confusion surrounding the construction of groups and the conditions for isomorphism. A group with two elements, such as G = {0, g} under addition, demonstrates that g is its own inverse, confirming its structure. The conclusion drawn is that any group of two elements can be mapped to another via a bijective homomorphism, reinforcing their isomorphic nature. Thus, the answer to the original question is that there is only one non-isomorphic group of two elements.
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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
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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