# Consequence of Lagrange's Theorem

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

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

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?

$$P:=\langle (12), (23)\rangle$$ 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*}
• 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 $abab$ – Jess Sep 18 at 4:05
• 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 $\langle a, b \rangle$. – sera Sep 18 at 4:09