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This is from Abstract Algebra, Dummit and Foote, pg 93.

Lagrange's thm consequence

For reference, this is how we know $|HK| = 4$: Proposition 13


My question is, how do we know $S_3 = \langle \: (12) , (23) \: \rangle $? What is it a consequence of?

I know it's not a subgroup because 4 doesn't divide $|S_3| = 6$. But couldn't {$(12) , (23)$} just be a set that is not a subgroup or group?

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$P:=\langle (12), (23)\rangle $ denotes the subgroup generated by $\{ (12), (23)\}$. That is, P is the smallest (w.r.t. set inclusion) subgroup of $S_3$ containing $\{ (12), (23)\}$. Since $P$ containing $(12)$ and $(13)$, $HK$ must be a subset of $P$. This is because $P$ is closed under group operation, $H$ is generated by $(12)$ and $K$ is generated by $(23)$. Thus $|P|\geq |HK|=4$. Since $P$ is a subgroup of $S_3$, its order divides $|S_3|=6$. Thus $|P|=6$ and $P=S_3$.

Note that $P$ contains all $6$ elements:

\begin{align*} & id = (1 2)(1 2) \\ & (1 2) \\ & (1 3) \\ & (1 2 3) = (1 3)(12) \\ & (1 3 2) = (1 2)(13) \\ & (2 3) = (1 2)(1 3)(1 2) \end{align*}

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  • $\begingroup$ Great explanation! I have a silly question about generator sets. If $\langle a,b \rangle$ is a generating set, then the group it generates is all posibilities of compositions since the composition is not necessarily commutative, correct? In other words, it is not just all $a^mb^l$ for integers l and m, but also all posible combinations such as $abab$ $\endgroup$ – Jess Sep 18 at 4:05
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    $\begingroup$ Yes. Any $x_1 x_2 \cdots x_n$ is an element of $\langle a, b \rangle$ if $x_i \in \{a, b, a^{-1}, b^{-1} \}$. For example, $a^m b^n a^k$ and $b^m a^n b^k a^{-1}bab^{-1} $ are elements of $\langle a, b \rangle$. $\endgroup$ – sera Sep 18 at 4:09

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