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Find the subgroup of $S_{3}$ generated by $\langle \bigl(\begin{smallmatrix} 1 & 2 & 3 & \\ 2 & 1 & 3 \end{smallmatrix} \!\!\bigr), \bigl(\begin{smallmatrix} 1 & 2 & 3 & \\ 1 & 3 & 2 \end{smallmatrix} \!\!\bigr) \rangle$

So in general I need to take all the combinations of the elements in the generating set (with their inverses) and see if I get all the elements of the group, or in this case $S_{3}$.

But what about the identity element? Do I need to "build it" from the elements of the generating set? or can I say that each the identity element is already included as "The trivial subgroup of any group is the subgroup $\{ e \}$ consisting of just the identity element."?

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  • $\begingroup$ Have you considered applying $(\begin{smallmatrix} 1 & 2 & 3 & \\ 2 & 1 & 3 \end{smallmatrix}) $ to itself? $\endgroup$
    – Meitar
    Commented Dec 15, 2016 at 16:44
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    $\begingroup$ @Meitar Yes, I saw that it the inverse of itself, but I was asking in general, do I always need to show that I can generate the identity element? $\endgroup$
    – gbox
    Commented Dec 15, 2016 at 16:45
  • $\begingroup$ What ervx stated pretty much takes care of that problem. In particular, an element being the inverse of itself also got generating the identity element covered... $\endgroup$
    – Meitar
    Commented Dec 15, 2016 at 16:48

4 Answers 4

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For any element $x$ in a finite group $G$, $x^{|x|}=e$ by definition, so $e$ will automatically be in any generating set. Here $|x|$ denotes the order of $x$ in $G$.

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    $\begingroup$ I think your aswer pretends not to see the real issue: in an infinte group $G$, a positive $\lvert x\rvert$ might not exist. $\endgroup$
    – user228113
    Commented Dec 15, 2016 at 16:50
  • $\begingroup$ That's true. But for any finite group, in particular $S_{3}$, I think the answer is fine. I included the word finite in my post now. $\endgroup$
    – ervx
    Commented Dec 15, 2016 at 16:52
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    $\begingroup$ This answer seems to miss that the subgroup generated by a set of elements always contains the identity, regardless of whether the elements have finite order. In the group $(\mathbb Z,+)$ the subgroup $\langle 1\rangle$ is the whole group, not the semigroup of positive integers. $\endgroup$ Commented Dec 15, 2016 at 16:58
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If $S$ is a subset of a group $G$, then the subgroup generated by $S$ is given by $$\langle S\rangle = \bigcap_{\substack{H\leq G\\S\subseteq H}}H.$$ That is, it is the intersection of every subgroup $H$ containing $S$. In particular, since the identity is in every subgroup, it is in the subgroup generated by any set. One can also use the definition that $\langle S\rangle$ is the elements of the form $s_1s_2\ldots s_i\in G$ where either $s_i$ or $s_i^{-1}$ is in $S$ for all $i$. Then, the empty product would be taken to give the identity, regardless of what $S$ is.

That is, subgroups are automatically closed under inverses and contain the identity, so you do not need to check these things. Note that you could speak of the subsemigroup generated by a set of elements - which would be exactly the set of products of the elements - in which case these are not guaranteed and would need to be check, but this is a distinct concept.

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So In general I need to take all the combinations of the elements in the generating set (with their inverses) and see if I get all the elements of the group, or in this case S3.

Let $a$ be one of your generating elements. Then, as you say, $a a^{-1}$ is in the generated subgroup, so yes, the identity element is automatically there. As to whether or not your teacher/grader will require you to note that fact every time, you'll have to ask them.

(There's a subtle question of "the subgroup generated by an empty set of generators" that I'm ignoring. You can probably ignore it too.)

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The subgroup generated by a set always contains the identity, by definition (it wouldn't be a subgroup if it didn't contain $e$). Milo Brandt's answer gives the formal definition.

Your intuition of $\left<\left(\begin{smallmatrix} 1 & 2 & 3 & \\ 2 & 1 & 3 \end{smallmatrix}\right),\left(\begin{smallmatrix} 1 & 2 & 3 & \\ 1 & 3 & 2 \end{smallmatrix}\right)\right>$ as "all combinations of the elements" is correct, but $e$ is a "trivial" combination. You can think of it as the zeroth power.

Put another way, $\left<\left(\begin{smallmatrix} 1 & 2 & 3 & \\ 2 & 1 & 3 \end{smallmatrix}\right),\left(\begin{smallmatrix} 1 & 2 & 3 & \\ 1 & 3 & 2 \end{smallmatrix}\right)\right>$ contains (among other things), every integer power of the element $a = \left(\begin{smallmatrix} 1 & 2 & 3 & \\ 2 & 1 & 3 \end{smallmatrix}\right)$. That is it contains $\ldots, a^{-2}, a^{-1}, a^0, a^1, a^2, \ldots$. But $a^0$ is $e$ for any group element, by definition. If that doesn't satisfy you, remember that it must contain both $a$ and $a^{-1}$, so it also contains the product $aa^{-1} =e$.

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