0
$\begingroup$

I have the following question :

Find sub group of $S_{10}$ that is isomorphic to $\mathbb{Z}_2 \oplus \mathbb{Z}_2$

What I did: I took a sub group that contains only $4$ elements and it has to be a disjoint so it would keep that property so I thought to take the following sub group (12),(34),(56),(78) and just define that $$f(12)=(0,0)$$ $$f(34)=(1,0)$$ $$f(56)=(0,1)$$ $$f(78)=(1,1)$$

I think that $f$ is isomorphic.

I wonder if my method is correct and if there's an easier to method (maybe using the first isomorphic theorem?) to solve such problems.

Thank you.

  • 0
    What you define is not a subgroup, it is not closed, $(12)(34)$ is not an element in your set with 4 elements. – 2017-01-27
  • 0
    @Mathematician42 What do you mean by closed? – 2017-01-27
  • 0
    @JaVaPG $\{(12),(34),(56),(78)\}$ is not a subgroup. – 2017-01-27
  • 1
    A group is closed under it's multiplication, that is if $a,b$ are two elements of a group $G$, then $ab$ must also lie in $G$. In your example or indeed the one above, this fails. – 2017-01-27

2 Answers 2

3

The Kleinian $4$-group $\mathbb{Z}_2\oplus \mathbb{Z}_2$ is a subgroup in any symmetric group $S_n$ for $n\ge 4$. We can take two generators consisting of disjoint transpositions, i.e., the group generated by $(ab)$ and $(cd)$, which is $\{id,(ab),(cd),(ab)(cd)\}$. In the case of $S_4$ there are many questions and answers on MSE for this, e.g. here, or here.

3

Hint: $\mathbb{Z}_2\oplus \mathbb{Z}_2$ is a subgroup of $S_4$.