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So the answer is

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1) How exactly is $<\pi>$ isomorphic to the other integer groups? I mean $\pi$ itself isn't even an integer.

2)What is exactly is the key saying for the single element sets? Are they trying to say they are isomorphic to themselves?

3) How exactly is $\{\mathbb{Z_6}, G \}$ and $\{ \mathbb{Z_2}, S_2\}$ isomorphic?

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up vote 2 down vote accepted

Hint: Consider the bijection $f:a^n\to n$.

To verify that $G$ and $\mathbb{Z}_6$ are indeed isomorphic let $$a=\begin{pmatrix} 1&2&3&4&5 \\ 3&5&4&1&2 \end{pmatrix}$$

Similarly for $\{S_2,\mathbb{Z}_2\}$ let $$a=\begin{pmatrix} 1&2 \\ 2&1 \end{pmatrix}$$ and isomorphism between the 2 groups should follow.

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Is $\mathbb{Z_6}$ considered to be a group under addition? – Hawk Nov 8 '12 at 5:43

1) Notice that $\langle \pi \rangle=\{\pi^n \mid n \in \mathbb Z\}$ and we also have $\pi^n\pi^m=\pi^{n+m}$ what group does this remind you of?

2) The single element sets are not isomorphic to any of the other groups. As an aside it's not particularly meaningful to say a group is isomorphic to itself. You generally speak of two groups that are distinct in some meaningful way as being isomorphic.

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So why do they include it in "as are..."? – Hawk Oct 24 '12 at 5:41

Jacob Schlather has already answered your first question, spelling out, you can consider the homomorphism $\varphi \colon \mathbb Z \to \langle \pi \rangle < \mathbb R^*$ given by $\varphi(n) = \pi^n$ for each $n \in \mathbb Z$.

About your second question, the exercise asks you to divide the given family of groups in isomorphism's class (i.e. it require to describe the equivalence classes for isomorphism relation inside the given family of groups).

Finally, addressing your third question, the group $G$ is generated by an element (it's a cyclic group) which can be written as product cyclic factors $(1\ 3\ 4)(2\ 5)$ so it's a cyclic-group generated by an element of order $6$ (luckily there aren't many groups of this type :) ). Similarly you have that the group $S_2$ is a group of order $2$ and also in this case there aren't many group of this type.

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