# *elements and generators of U(14)

• MHB
• karush

#### karush

Gold Member
MHB
elements and generators of U(14)

\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,
&\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{\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

Hi karush,

Here are a few comments on what you've written:

(b) Your set for $U(14)$ needs revision. Try examining it again and seeing what needs to be changed.

(f) The subgroup generated by 5 needs revision as well. When looking at a subgroup generated by an element, one big hint that something is not quite right is when the identity is not present in the subgroup.

(g) You're correct that $U(14)$ has order 6, which conflicts with part (b) because you have 11 elements listed.

(h) It's not quite clear what you meant when you say $U(14)$ has subgroups, because every group always has at least "two" subgroups - namely the group itself and the subgroup consisting only of the identity. Note: "two" could be one in the trivial case that the group consists only of the identity.

I hope this helps get things going in the right direction.

mahalo