Chapter 13

If \(G\) has order \(n\), then \(x^{n}=e\) for every \(x\) in \(G\).

The subgroup \(\langle 3\rangle\) of \(\mathbb{Z}\).

The order of \(H \cap K\) is a common divisor of the order of \(H\) and the order of \(K\).

Let \(G\) have order \(p q\), where \(p\) and \(q\) are primes. Either \(G\) is cyclic, or every element \(x \neq e\) in \(G\) has order \(p\) or \(q\)

Let \(G\) have order 4 . Either \(G\) is cyclic, or every element of \(G\) is its own inverse. Conclude that every group of order 4 is abelian.

if \(b a=a b^{2}\), prove that \(b a^{2}=a^{2} b^{4}\), and conclude that \(b=b^{4} .\) This is impossible cause \(b\) has order 5, hence \(b a \neq a b^{2}\), (HINT: The equation \(b a=a b^{2}\) tells us that may move a factor \(a\) from the right to the left of a factor \(b\), but in so doing, we ist square \(b .\) To prove an equation such as the preceding one, move all factors \(a\) the left of all factors \(b\).)

If \(G\) has an element of order \(p\) and an element of order \(q\), where \(p\) and \(q\) are distinct primes, then the order of \(G\) is a multiple of \(p q\).

Suppose \(H\) has index \(p\) and \(K\) has index \(q\), where \(p\) and \(q\) are distinct primes. hen the index of \(H \cap K\) is a multiple of \(p q\).

The number of right cosets of \(H\) is equal to the number of left cosets of \(H\).

If \(G\) is an abelian group of order \(n\), and \(m\) is an integer such that \(m\) and \(n\) are atively prime, then the function \(f(x)=x^{m}\) is an automorphism of \(G\).

For any positive integer \(m\), what is the index of \(\langle m\rangle\) in \(\mathbb{Z}\) ?