- #1

karush

Gold Member

MHB

- 3,269

- 5

\begin{align*}\displaystyle

&\text{(a)the identity is } \color{red}{1} \\

&\text{(b) U(14) is the set } \color{red}{\{1,3,4,5,6,8,9,10,11,12,13\}}\\

&\text{(c) |1|}={\color{red}{1}} \text{ since }1.1 \equiv 12^1\\

&(d) |13|={\color{red}{2}}

\text{ since }(13)^1=13\ne 1,

(13)^2 \equiv _{14} ^{\quad(-1)^2}=1\\

&\text{(e) the inverse of 13 is } {\color{red}{13}}\\

&\quad\text{ Since} 13^2 = 1 \mod 14, 13 \text{is its own inverse.}\\

&\text{(f) the generator of this group is }\\

&\quad\text{The subgroup generated by}\\

&\quad{\color{red}{<3>}}=\{3^k| k \in \Bbb{Z}\} = \{3,9,13,11,5,1\}\\

&\quad{\color{red}{<5>}}=\{5^k| k \in \Bbb{Z}\} = \{5,11 13,9,3\}\\

&(g) Abelian/non-Abelian? \\

&\quad\text{Abelian group of order } \color{red}{6}\\

&\text{(h) U(14) has subgroups.}\\

&\quad\textit{<11>}=\{11^k|k\in \Bbb{Z}\} = \{ 11, 9, 1 \} 6 \ne U(14)

\end{align*}

hopefully

(d) (h) was guesstimates? others maybe