AATA Lagrange’s Theorem

Section 6.2 Lagrange’s Theorem

Proposition 6.9.

Let \(H\) be a subgroup of \(G\) with \(g \in G\) and define a map \(\phi:H \rightarrow gH\) by \(\phi(h) = gh\text{.}\) The map \(\phi\) is bijective; hence, the number of elements in \(H\) is the same as the number of elements in \(gH\text{.}\)

Proof.

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 \(\phi(h) = gh\text{.}\)

Theorem 6.10. Lagrange.

Let \(G\) be a finite group and let \(H\) be a subgroup of \(G\text{.}\) Then \(|G|/|H| = [G : H]\) is the number of distinct left cosets of \(H\) in \(G\text{.}\) In particular, the number of elements in \(H\) must divide the number of elements in \(G\text{.}\)

Proof.

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

Corollary 6.11.

Suppose that \(G\) is a finite group and \(g \in G\text{.}\) Then the order of \(g\) must divide the number of elements in \(G\text{.}\)

Corollary 6.12.

Let \(|G| = p\) with \(p\) a prime number. Then \(G\) is cyclic and any \(g \in G\) such that \(g \neq e\) is a generator.

Proof.

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 \({\mathbb Z}_p\text{.}\)

Corollary 6.13.

Let \(H\) and \(K\) be subgroups of a finite group \(G\) such that \(G \supset H \supset K\text{.}\) Then

\begin{equation*} [G:K] = [G:H][H:K]\text{.} \end{equation*}

Proof.

Observe that

\begin{equation*} [G:K] = \frac{|G|}{|K|} = \frac{|G|}{|H|} \cdot \frac{|H|}{|K|} = [G:H][H:K]\text{.} \end{equation*}

Remark 6.14. The converse of Lagrange’s Theorem is false.

The group \(A_4\) has order \(12\text{;}\) however, it can be shown that it does not possess a subgroup of order \(6\text{.}\) According to Lagrange’s Theorem, subgroups of a group of order \(12\) can have orders of either \(1\text{,}\) \(2\text{,}\) \(3\text{,}\) \(4\text{,}\) or \(6\text{.}\) However, we are not guaranteed that subgroups of every possible order exist. To prove that \(A_4\) has no subgroup of order \(6\text{,}\) 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.

Proposition 6.15.

The group \(A_4\) has no subgroup of order \(6\text{.}\)

Proof.

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 \((1 \, 2 \, 3)\) is in \(H\text{.}\) Then \((1 \, 2 \, 3)^{-1} = (1 \, 3 \, 2)\) 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*} (1 \, 2 \, 4)(1 \, 2 \, 3)(1 \, 2 \, 4)^{-1} & = (1 \, 2 \, 4)(1 \, 2 \, 3)(1 \, 4 \, 2) = (2 \, 4 \, 3)\\ (2 \, 4 \, 3)(1 \, 2 \, 3)(2 \, 4 \, 3)^{-1} & = (2 \, 4 \, 3)(1 \, 2 \, 3)(2 \, 3 \, 4) = (1 \, 4 \, 2) \end{align*}

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

\begin{equation*} (1), (1 \, 2 \, 3), (1 \, 3 \, 2), (2 \, 4 \, 3), (2 \, 4 \, 3)^{-1} = (2 \, 3 \, 4), (1 \, 4 \, 2), (1 \, 4 \, 2)^{-1} = (1 \, 2 \, 4)\text{.} \end{equation*}

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

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

Theorem 6.16.

Two cycles \(\tau\) and \(\mu\) in \(S_n\) have the same length if and only if there exists a \(\sigma \in S_n\) such that \(\mu = \sigma \tau \sigma^{-1}\text{.}\)

Proof.

Suppose that

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

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

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