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I'm trying to check if the groups $A_4$ and $S_3 \times \Bbb Z_2$ are or not isomorphic. How can I check if they are? I'm trying to understand how can I generally prove an isomorphism with this kind of groups. Any help?

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  • $\begingroup$ Tip: You can use \times to generate a product symbol in LaTeX: $\times$ $\endgroup$ – Wojowu Mar 21 at 21:18
  • $\begingroup$ @Wojowu fixed. thanks $\endgroup$ – Jack Mar 21 at 21:28
  • $\begingroup$ The usual way to prove two groups are isomophic is to produce an explicit isomorphism. As noted in the answer, many group properties can be used to show that two groups are not isomorphic. In this case what I think you'll find useful is counting elements of order $6$. $\endgroup$ – Robert Shore Mar 21 at 23:24
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When proving that two groups aren't isomorphic, you need to find some property which shows they are different. The easiest one is the number of elements. Unfortunately they both have $12$ elements, so we're out of luck there.

The next step in the same direction is to look at the orders of elements: How many elements do each of the two groups have of order $2$? How many elements of order $3$? $4$? $6$? $12$? Is there any of those for which our groups are different? If yes, then the groups cannot be isomorphic.

As you keep learning about group theory, you'll learn about more things you can use to to differentiate groups: The structure of subgroups, of normal subgroups, the center, the derived subgroup, and so on.

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  • $\begingroup$ Thank you. Just to understand... To be isomorphic the two groups must have the same number of elements with a certain order. Is that correct? Also, how do I calculate the order of an element of $S_3 \times \Bbb Z_2$ For example, let's take $((1 2), 1) \in S_3 \times \Bbb Z_2$ How can I generally calculate the order of this kind of pair? Is it 2 since their order of the permutation is 2 and $1^2=1$? Or it is not the correct method to calculate their order? $\endgroup$ – Jack Mar 22 at 9:47
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    $\begingroup$ @Jack Yes, for two groups to be isomorphic the number of elements of any given order must be the same (and this isn't that hard to prove, since if $f$ is an isomorphism, $a$ and $f(a)$ have the same order, and $f$ is a bijection). And you almost got the order. Remember that in $\Bbb Z_2$, the operation is addition. So when we say $1^2$, we mean $1+1$, and the identity element is $0$. So your element has order $2$ because $(12)^2 = e$ and $1^2 = 0$. The element $((123), 1)$ has order $6$. In general, in a prodict $G\times H$, the order of $(g, h)$ is the least common multiple of $|g|$ and $|h|$ $\endgroup$ – Arthur Mar 22 at 9:53
  • $\begingroup$ (And yes, mixing multiplicative and additive group notations gets confusing.) $\endgroup$ – Arthur Mar 22 at 9:53
  • $\begingroup$ Ok, I tried to find the order of their elements. I found that has only elements of order 3 and 2 (and, of course, the identity). $S_3 \times \Bbb Z_2$ has an element of order 1 (???) which is (identity, 0), order 2, 3, 6. So basically there isn't an isomorphism (even just because there aren't elements of order 6 in A4). Please, correct me if I'm wrong. $\endgroup$ – Jack Mar 22 at 10:49
  • $\begingroup$ @Jack The absence of order $6$ elements in $A_4$ while $S_3\times \Bbb Z_2$ has two of them is exactly the kind of thing that lets you conclude that there cannot be an isomorphism between them. And yes, the identity element of a group always has order $1$, and it is always the only one that does. The identity of $S_3\times\Bbb Z_2$ is $(e, 0)$ (where $e$ is the trivial permutation). $\endgroup$ – Arthur Mar 22 at 11:37
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Here's one that kind of jumps out at me for this pair of groups. $S_3 \times \Bbb Z_2$ has a non-trivial center; in other words, it has a non-identity element that commutes with every element of the group. (Can you see what it is?) $A_4$ has no such element.

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