U(23)

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Group: U(23)
Group Description: The set of integers less than 23 and relatively prime to 23 with multiplication modulo 23.

Group Definition: {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22}.
For the purpose of this project, we will define $Z$ = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22}.

Prove that your structure is in fact a group

Now that we know what U(23) is, we must prove that it is a group. To prove that U(23) is a group, we must verify that the potential group satisfies the three Group Axioms.

Group Axioms:
$\mathcal{G}_1$: $\textbf{associativity of *}$ For all $\textit{a, b, c,} \in \textit{G}$, we have
$(\textit{a}*\textit{b})*\textit{c}=\texit{a}*(\textit{b}*\textit{c}).$
$\mathcal{G}_2$: $\textbf{identity element} \textit {e} for *.$ There is an element $\textit{e}$ in $\textit{G}$ such that for all x $\in \textit{G}$.
$\textit{e} *\textit{x} = \textit {x} * \textit {e}=\textit {x}$.
$\mathcal{G}_3$: $\textbf{inverse} \textit{a} ^\prime of \textit{a}$. Corresponding to each $\textit{a}\in\textit{G}$, there is an element $\textit {a} ^\prime$ in$\textit{G}$ such that
$\textit{a}*\textit{a} ^\prime=\textit{a} ^\prime*\textit{a}=\textit{e}$.

Proof

$\mathcal{G}_1$: $Z$ is associative because it is based on $\Bbb{Z}$ and $\Bbb{Z}$ is associative under multiplication so $\mathcal{G}_1$ is satisfied.

$\mathcal{G}_2$: $Z$ must have an identity under multiplication modulo 23 for $Z$. Let the $id = e$. Then $mod(a \times e,23) = a$ where $a$ is an element of $Z$. The identity for multiplication is 1 so we will check to see if it satisfies the given equation.

$mod(a \times 1,23) = a$.

Therefore, 1 is the identity for $Z$ and $\mathcal{G}_2$ is satisfied.

$\mathcal{G}_3$: For every element $a \in Z$, there must exist a $b \in Z$ such that $mod(a \times b,23) = e$ where $e = 1$.
Since we have a finite list of elements, we will compute each inverse as follows.

$mod(1 \times 1,23) = 1$
Therefore 1 is the inverse for itself.
$mod(2 \times 12,23) = 1$
Therefore 2 and 12 are inverses of each other.
$mod(3 \times 8,23) = 1$
Therefore 3 and 8 are inverses of each other.
$mod(4 \times 6,23) = 1$
Therefore 4 and 6 are inverses of each other.
$mod(5 \times 14,23) = 1$
Therefore 5 and 14 are inverses of each other.
$mod(7 \times 10,23) = 1$
Therefore 7 and 10 are inverses of each other.
$mod(9 \times 18,23) = 1$
Therefore 9 and 18 are inverses of each other.
$mod(11 \times 21,23) = 1$
Therefore 11 and 21 are inverses of each other.
$mod(13 \times 16,23) = 1$
Therefore 13 and 16 are inverses of each other.
$mod(15 \times 20,23) = 1$
Therefore 15 and 20 are inverses of each other.
$mod(17 \times 19,23) = 1$
Therefore 17 and 19 are inverses of each other.
$mod(22 \times 22,23) = 1$
Therefore 22 is the inverse for itself.

Thus, we see that each element in $Z$ has an inverse also in $Z$ and $\mathcal{G}_3$ is satisfied.

Therefore, we can conclude that $Z$ is a group because the 3 axioms are satisfied.

Find some (or all) non-trivial, proper subgroups of your group

Non-trivial, proper subgroups
After many calculations, attempting to find a proper subgroup of $Z$ to find none leaves me to believe that there are no proper subgroups that are non-trivial. Every attempt would lead me back to $Z$ or leave me with a set that is not closed which means it cannot be a subgroup.
page revision: 24, last edited: 02 Nov 2010 06:52