Let \(G\) be a finite group. Then the order of an element \(g\) is the smallest number \(n\) such that \(g^n=e\). Show that the order of \(g\in G\) is finite group.
Proof: Since \(G\) is a finite group, then \(|G|<\infty\). Let

We have \(S\subseteq G\), so \(S\) is finite. Thus there must exists \(p,q\in\mathbb N,p<q\), such that \(g^p=g^q\Rightarrow g^{q-p}=e\), where \(e\) is the identity of \(G\). Therefore, we have \(n\le q-p\) by the definition of the order of an element \(g\). Since \(q-p\) is finite, the order of \(g\) is finite. #

Let \(G\) be a group with order \(|G|=n\). \(S\) is a subset of \(G\), with \(|S|>\frac{n}{2}\). Show that for any \(g\in G\), there exists \(a,b\in S\) such that \(g=ab\).
Proof: For any \(g\in G\), construct a map

Note that for any \(h_1,h_2\in G\), let \(f(h_1)=f(h_2)\), we have

Thus \(f\) is injective. On the other hand, for any \(h\in G\), \(\exists gh^{-1}\), such that \(f(gh^{-1})=(gh^{-1})^{-1}g=hg^{-1}g=h\). So \(f\) is surjective. Therefore, \(f\) is bijective.

Suppose that \(f(S)\cap S=\varnothing\), then \(|f(S)\cup S|=|f(S)|+|S|=2|S|\). However, \(f(S)\cup S\subseteq G\), it follows that \(2|S|=|f(S)\cup S|\le |G|=n\Rightarrow |S|\le \frac{n}{2}\). \(\to\leftarrow\).

Therefore, \(f(S)\cap S\neq\varnothing\). Let \(b\in f(S)\cap S\). Clearly, \(b\in S\). Then exists \(a\in S\), such that \(f(a)=b\Rightarrow a^{-1}g=b\), i.e. \(g=ab\). #

Let \(a,b\) be two elements of a group \(G\), and \(aba=ba^2b,~a^3=1,~b^{2n-1}\). Then \(b=1\).
Proof: \(aba=ba^2b\Rightarrow aba^3=ba^2ba^2\Rightarrow ab=ba^2ba^2\Rightarrow ab^2=ba^2ba^2b=ba^2aba=b^2a\). i.e. The element \(a\) commutes with \(b^2\). Thus

Therefore, \(ab=ab^{2n-1}b=b^{2n-1}ba=ba\Rightarrow ba^2b=aba=ba^2\Rightarrow(ba^2)^{-1}(ba^2)b=(ba^2)^{-1}(ba^2)=1\), i.e. \(b=1\). #