Table of Content
Let \(H\) be a subgroup of a finite group \(G\). Then the order of \(H\) divides the order of \(G\)。
Proof
Since \(G\) is equal to the disjoint union of left cosets of \(H\) and every left coset has the same cardinality as \(H\), we have
\[\vert G \vert = \vert H \vert \times (\text{# of left cosets}).\]This proves the theorem.
看看這個 theorem。
The order of an element \(a\) of a finite group \(G\) divides the order of \(G\).
Proof
Let \(H=\langle a \rangle\), and apply Lagrange’s Thm. ◼
於是 \(g\) 是 \(G\) 的 generator 的充分必要條件就是：For all \(d\) divides \(n\) and \(d\not = n\),
\[g^d \not = 1.\]Every group \(G\) of prime order is cyclic.
Proof
Let \(g \in G,\ g \not = e\). We have \(\vert \langle g \rangle \vert\) divides \(\vert G \vert\). Since \(\vert G \vert\) is prime, \(\vert \langle g \rangle \vert\) is either \(1\) or \(\vert G \vert\). The former case is impossible since \(g \not = e\). Then \(\vert \langle g \rangle \vert = \vert G \vert\) implies \(\langle g \rangle = G\), i.e. \(G\) is cyclic.
Let \(H\) be a subgroup of a group \(G\). The number of left cosets of \(H\) is the index of \(H\) in \(G\), which is denoted by \([G:H]\).
Suppose \(K \le H \le G\), and \([G:H]\) and \([H:K]\) are finite. Then \([G:K]=[G:H][H:K]\).
依定義展開就可證明。