Partition groups into subcollection

[kathrynag]

Homework Statement

Partition the following collection of groups into subcollections of isomorphic groups. a * superscript means all nonzero elements of the set.
integers under addition
$$S_{2}$$
$$S_{6}$$
$$integer_{6}$$
$$integer_{2}$$
$$real^{*}$$ under multiplication
$$real^{+}$$ under multiplication
$$rational^{*}$$ under multiplication
$$complex^{*}$$ under multiplication
17(integer) under addition
rational under addition
The subgroup (pi) of $$real^{*}$$ under multiplication
3(integer) under multiplication
real under addition
The subgroup G of $$S_{5}$$ generated by :
top row:(1 2 3 4 5)
bottom row (3 5 4 1 2)

Homework Equations

The Attempt at a Solution

i don't understand the question and what this has to do with the permutation chapter in the book.

 

[kathrynag]

Does this make any sense with the permutation chapter

 

[kathrynag]

I'm still clueless on this one. If anyone has any ideas, it would really help. I'm racking my brain and not coming up with anything.

 

[yyat]

"Partition the following collection of groups into subcollections of isomorphic groups" is just a fancy way of asking which of these groups are isomorphic.

To determine if two groups are isomorphic you can use cardinality (number of elements) as a first hint, if they don't have the same cardinality, they can't be isomorphic. To prove that two groups actually are isomorphic, you need to construct an isomorphism (the exponential function should be helpful in some cases).

 

[kathrynag]

so, like s6 and integer6 would have same cardinality?

 

[yyat]

so, like s6 and integer6 would have same cardinality?

No, $$S_6$$ is a http://en.wikipedia.org/wiki/Symmetric_group" and has 6!=720 elements, the cyclic group $$\mathbb{Z}_6$$ on the other hand has only 6 elements.

But try $$S_2$$ and $$\mathbb{Z}_2$$.

 

[kathrynag]

Oh, I see since S2 has 2! elements and Z2 has 2 elements

 

[kathrynag]

Ok, I have an idea.
I have s2 and z2 as one collection
I have z,3z,17z as one collection
My problem is what to do with the reals, rationals and complex?

 

[yyat]

Ok, I have an idea.
I have s2 and z2 as one collection
I have z,3z,17z as one collection

What is 3z? Is it $$\mathbb{Z}_3$$?

My problem is what to do with the reals, rationals and complex?

Think again about cardinality: The rationals are countable, the reals are uncountable. Also consider my hint about the exponential function (Post #4), it relates addition and multiplication.

 

[kathrynag]

3z is the set of integers Z times 3.
Ok reals are uncountable. is this true for complex, too?
I'm just having trouble with the exponential; function and how to use it.
Are real* and complex* under multiplication related?

 

[e(ho0n3]

The reals are a subset of the complexes and if the reals are uncountable, then ...

Of course, that doesn't mean they both have the same cardinality.

A further hint about the exponential function: exp(x + y) = exp(x) * exp(y)

 

[kathrynag]

then complex are uncountable
ok, ao does that hint say something like reals under addition and real under multiplications are a collection?

 

[e(ho0n3]

then complex are uncountable

Good.

ok, ao does that hint say something like reals under addition and real under multiplications are a collection?

You meant to say isomorphic instead of collection, I hope. Can you prove, use exp, that they are isomorphic?

 

[kathrynag]

yeah, I meant to say isomorphic.
let x and y be 2 real numbers under addition
Then exp(x+y)=exp(x)*exp(y)

 

[kathrynag]

Ok still having trouble.
I'm saying R^+, R is one isomorphism
C*,R* is one isomorphism
Q,Q* is one isomorphism

 

[kathrynag]

Is this correct?