Proposition6.9

Let $H$ be a subgroup of $G$ with $g \in G$ and define a map $\phi:H \rightarrow gH$ by $\phi(h) = gh\text{.}$ 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 $\phi(h) = gh\text{.}$

Theorem6.10Lagrange

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, $|G| = [G : H] |H|\text{.}$

Corollary6.11

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

Corollary6.12

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 $G\text{.}$

Corollary6.13

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

Proof

Observe that

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.

Proposition6.15

The group $A_4$ has no subgroup of order 6.

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 $(123)$ 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

we can conclude that $H$ must have at least seven elements

Therefore, $A_4$ has no subgroup of order 6.

Theorem6.16

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

Define $\sigma$ to be the permutation

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,

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