8 Let \(G\) be any multiplicative group. \(H\) is a subset of \(G\). \(H\) consists of all elements \(h\) such that \(h g = g h\) for every element \(g\) in \(G\).
- Prove that \(H\) is a subgroup of \(G\).
Now consider the case where \(G\) is given by the following table:
| \(e\) | \(p\) | \(q\) | \(r\) | \(s\) | \(t\) |
| \(e\) | \(e\) | \(p\) | \(q\) | \(r\) | \(s\) | \(t\) |
| \(p\) | \(p\) | \(q\) | \(e\) | \(s\) | \(t\) | \(r\) |
| \(q\) | \(q\) | \(e\) | \(p\) | \(t\) | \(r\) | \(s\) |
| \(r\) | \(r\) | \(t\) | \(s\) | \(e\) | \(q\) | \(p\) |
| \(s\) | \(s\) | \(r\) | \(t\) | \(p\) | \(e\) | \(q\) |
| \(t\) | \(t\) | \(s\) | \(r\) | \(q\) | \(p\) | \(e\) |
- Show that \(H\) consists of just the identity element.