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Section6.2Lagrange's Theorem

We first show that the map \(\phi\) is one-to-one. Suppose that \(\phi(h_1) = \phi(h_2)\) for elements \(h_1, h_2 \in H\text{.}\) We must show that \(h_1 = h_2\text{,}\) but \(\phi(h_1) = gh_1\) and \(\phi(h_2) = gh_2\text{.}\) So \(gh_1 = gh_2\text{,}\) and by left cancellation \(h_1= h_2\text{.}\) To show that \(\phi\) is onto is easy. By definition every element of \(gH\) is of the form \(gh\) for some \(h \in H\) and \(\phi(h) = gh\text{.}\)

The group \(G\) is partitioned into \([G : H]\) distinct left cosets. Each left coset has \(|H|\) elements; therefore, \(|G| = [G : H] |H|\text{.}\)

Let \(g\) be in \(G\) such that \(g \neq e\text{.}\) Then by Corollary 6.11, the order of \(g\) must divide the order of the group. Since \(|\langle g \rangle| \gt 1\text{,}\) it must be \(p\text{.}\) Hence, \(g\) generates \(G\text{.}\)

Corollary 6.12 suggests that groups of prime order \(p\) must somehow look like \({\mathbb Z}_p\text{.}\)

Observe that

\begin{equation*} [G:K] = \frac{|G|}{|K|} = \frac{|G|}{|H|} \cdot \frac{|H|}{|K|} = [G:H][H:K]. \end{equation*}
Remark6.14The converse of Lagrange's Theorem is false

The group \(A_4\) has order 12; however, it can be shown that it does not possess a subgroup of order 6. According to Lagrange's Theorem, subgroups of a group of order 12 can have orders of either 1, 2, 3, 4, or 6. However, we are not guaranteed that subgroups of every possible order exist. To prove that \(A_4\) has no subgroup of order 6, we will assume that it does have such a subgroup \(H\) and show that a contradiction must occur. Since \(A_4\) contains eight 3-cycles, we know that \(H\) must contain a 3-cycle. We will show that if \(H\) contains one 3-cycle, then it must contain more than 6 elements.

Since \([A_4 : H] = 2\text{,}\) there are only two cosets of \(H\) in \(A_4\text{.}\) Inasmuch as one of the cosets is \(H\) itself, right and left cosets must coincide; therefore, \(gH = Hg\) or \(g H g^{-1} = H\) for every \(g \in A_4\text{.}\) Since there are eight 3-cycles in \(A_4\text{,}\) at least one 3-cycle must be in \(H\text{.}\) Without loss of generality, assume that \((123)\) is in \(H\text{.}\) Then \((123)^{-1} = (132)\) must also be in \(H\text{.}\) Since \(g h g^{-1} \in H\) for all \(g \in A_4\) and all \(h \in H\) and

\begin{align*} (124)(123)(124)^{-1} & = (124)(123)(142) = (243)\\ (243)(123)(243)^{-1} & = (243)(123)(234) = (142) \end{align*}

we can conclude that \(H\) must have at least seven elements

\begin{equation*} (1), (123), (132), (243), (243)^{-1} = (234), (142), (142)^{-1} = (124). \end{equation*}

Therefore, \(A_4\) has no subgroup of order 6.

In fact, we can say more about when two cycles have the same length.

Suppose that

\begin{align*} \tau & = (a_1, a_2, \ldots, a_k )\\ \mu & = (b_1, b_2, \ldots, b_k ). \end{align*}

Define \(\sigma\) to be the permutation

\begin{align*} \sigma( a_1 ) & = b_1\\ \sigma( a_2 ) & = b_2\\ & \vdots \\ \sigma( a_k ) & = b_k. \end{align*}

Then \(\mu = \sigma \tau \sigma^{-1}\text{.}\)

Conversely, suppose that \(\tau = (a_1, a_2, \ldots, a_k )\) is a \(k\)-cycle and \(\sigma \in S_n\text{.}\) If \(\sigma( a_i ) = b\) and \(\sigma( a_{(i \bmod k) + 1}) = b'\text{,}\) then \(\mu( b) = b'\text{.}\) Hence,

\begin{equation*} \mu = ( \sigma(a_1), \sigma(a_2), \ldots, \sigma(a_k) ). \end{equation*}

Since \(\sigma\) is one-to-one and onto, \(\mu\) is a cycle of the same length as \(\tau\text{.}\)