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 $gH\\text{.}$
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 $\\phi(h) = gh\\text{.}$
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 $G\\text{.}$
The group $G$ is partitioned into $[G : H]$ distinct left cosets. Each left coset has $|H|$ elements; therefore, $|G| = [G : H] |H|\\text{.}$
Suppose that $G$ is a finite group and $g \\in G\\text{.}$ Then the order of $g$ must divide the number of elements in $G\\text{.}$
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.
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 ${\\mathbb Z}_p\\text{.}$
Let $H$ and $K$ be subgroups of a finite group $G$ such that $G \\supset H \\supset K\\text{.}$ Then
Observe that
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.
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.
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 $\\mu = \\sigma \\tau \\sigma^{-1}\\text{.}$
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 $\\tau\\text{.}$