Show whether a subset of $S_3$ is or is not a subgroup Show whether the subset $\{(1),(12),(23),(13)\}$ of $S_3$ is or is not a subgroup.
Not too sure about this, pretty new to group theory in general. I get that $S_3$ is the symmetric group and I know what a subset is, but do I just have to show that the elements of the subset can be multiplied together and remain in $S_3$?
 A: The given subset is not a subgroup of $S_3$. For one, its order does not divide that of $S_3$, which is a necessary condition for a subgroup: $4\nmid6$. In addition, composing $(12)$ and $(23)$ produces a permutation outside the subset, $(231)$, violating the closure property of all groups.
A: If $H$ is a subset of a group $G$ then $H$ will be a subgroup of $G$ if and only if $H$ is not empty and for all $a,b\in H$ we have $ab^{-1}\in H$. 
This is evidently necessary. Further note that - if the condition is satisfied so that $H$ contains some element $c$ - we find that: 


*

*$e=cc^{-1}\in H$.

*$b^{-1}=eb^{-1}\in H$ for every $b\in H$

*$ab=a(b^{-1})^{-1}\in H$ for every $a\in H$ and every $b\in H$


So $H$ contains the identity, has inverses and is closed under multiplication, hence is a subgroup.

If $H$ is a subset of a finite group $G$ then this condition can be "weakened": $H$ is a subgroup of $G$ if and only if $H$ is not empty and for all $a,b\in H$ we have $ab\in H$.
Again it is evident that the condition is necessary. Further if $a\in H$ and the condition is satisfied then $a^k\in H$ for every positive integer $k$. Because $G$ is finite we must have $a^n=e$ for some positive integer $n$ so that $e=a^n\in H$. This because the map $\mathbb N\to G$ prescribed by $n\mapsto a^n$ cannot be injective and $a^k=a^l$ with $k<l$ leads to $a^{l-k}=e$. Then $e=a^{l-k}\in H$ and also $a^{-1}=a^{l-k-1}\in H$. So if $a,b\in H$ then $a,b^{-1}\in H$ and - if the condition is satisfied - $ab^{-1}\in H$. This shows that the original condition is satisfied.
