##### Proposition6.9

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

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

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

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

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.

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

Let \(H\) and \(K\) be subgroups of a finite group \(G\) such that \(G \supset H \supset K\). Then \begin{equation*}[G:K] = [G:H][H:K].\end{equation*}

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.

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

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

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