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Show that $S_2$ is isomorphic to $\mathbb{Z}_2$

I know that $S_2$ has two elements $\sigma_1$ and $\sigma_2$...and $\mathbb{Z}_2$ has two elements $0$ and $1$. But isomorphism is a bijective function, right? So doesn't the question need to give me a function so I can prove that there is an isomorphism? In other words, how can I prove that something is isomorphic to something else when there when there is no function to begin with?

Thanks in advance.

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Try to write down the tables of two groups containing the compositions of their elements, and examine them. – Babak S. Feb 7 '13 at 12:23
up vote 2 down vote accepted

I assume that $S_2=\langle\sigma_1,\sigma_2\rangle$ wherein $\sigma_1=e_{S_2}$ and $\sigma_2=(1, 2)$. So we have the following table illustrating the whole group.

enter image description here

And for $\mathbb Z_2$ we can establish an small table as $S_2 $:

enter image description here

By a simple examining between two tables, you can get that whenever $\sigma_1$ plays a rule as $S_2$'s identity; $0$ does the same in another group. Both are finite, and so there is a natural bijection function between them. In fact: $$S_2\cong\mathbb N_2\cong\mathbb Z_2$$

You have to show that this function is a homomorphism too. By that, you will show that the map is an isomorphism. For example: $$f:S_2\longrightarrow \mathbb Z_2\\\sigma_1\to 0\\\sigma_2\to1$$

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Thanks a lot. So I say f( (1)(2)(1 2) = f((1 2)) = 1 and f((1)(2))f((1 2))=0*1=1...I need to check this for all possibilities, right? After checking those...the function is obviously a one-to-one do I have to "prove" that too or can I just leave it since it is obvious? Also, I could have defined a function as f: $Z_2 \rightarrow S_2$, right? So there isn't just one choice for the just needs to be an isomorphism, right? – user58289 Feb 7 '13 at 12:57
Yes, all you said is right but just show that, for example, the function I noted is a group homomorphism. In fact, show that $$f(x*_{S_{2}}y)=f(x)*_{\mathbb Z_2}f(y),~~~x,y\in S_2$$ – Babak S. Feb 7 '13 at 13:01
But that's what I already showed...I said f( (1)(2)(1 2) = f((1 2)) = 1 and f((1)(2))f((1 2))=0*1=1. So f((1)(2)(1 2))=f((1)(2))f((1 2))=1. I just need to show this for the other possibilites and then I'm done, right? – user58289 Feb 7 '13 at 13:11
@Artus: You did it. ;-) – Babak S. Feb 7 '13 at 13:12
I like the graphics/tables! +1 – amWhy Feb 8 '13 at 0:04

You have to construct such a function, that's what's required from you.

Actually, there isn't much choice. There are only two elements in each group, and you should know that the identity has to go to the identity...

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You have to guess what the function is. In the case of a morphisms, the neutral element of the first group has to be mapped to the neutral element of the second group. As an isomorphism is in particular a bijection, the transposition $(1 \: 2)$ has to be mapped to the the class of $1$ in $\Bbb Z_2$.

So we have an unique candidate to be an isomorphism. Now we need to check it indeed fulfills the conditions.

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Keep in mind that any group $G$ of prime order $p$ is isomorphic to $\mathbb{Z}_p$. (Try to prove it by showing that an element of order $p$ generates $G$)

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When you are asked to prove that two groups are isomorphic, you generally don't get any function. Sometimes you have to find it yourself, and sometimes you can use some theorems to prove they're isomorphic without knowing a particular isomorphism, in this case, the isomorphism is trivial. If we look at $S_2$ as the permutations of a set of two elements, then there can be only two, the identity and one that swaps them, lets call them $id$ and $g$, so $$S_2=\lbrace id,g\rbrace$$, and $$\mathbb{Z_2}=\lbrace 0,1\rbrace$$ Can't you think of any function between them, it's reasonable to asign the identity of each group to the identity of the other, and the other element to the one left, so we define: $$f:S_2\longrightarrow \mathbb{Z}_2$$ such that: $$f(id)=0$$ $$f(g)=1$$ I leave it to you to prove that $f$ is an isomorphism.

Another way: We could use elementary facts about groups: all groups have an element of order 1 that is called the identity; if a group has two elements, it has the identity $1$ and another element, let's call it $a$. From here, we prove the group is abelian: (trivial) $a1=1a=a$. So every two element group is abelian, and by the theorem of structure of finite abelian groups they all must be isomorphic to $\mathbb{Z}_2$, so we have proved that all groups with two elements are isomorphic to each other.

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