Hw10 Problem 5

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Problem 5

Suppose that $\phi: \mathbb{Z_{40}^*} \rightarrow \mathbb{\mathbb{Z_{40}^*}}$ is a homomorphism. (Recall, $\mathbb{Z_{40}^*}$ is the group of all elements of $\mathbb{Z_{40}}$ that have a multiplicative inverse with operation $\cdot$ modulo $40$.) If $Ker \phi = \{1,9, 17,33\}$ and $\phi(11)=11$, find all elements of $\mathbb{Z_{40}^*}$ that map to $11$.

Solution

By Theorem 13.15, if $\phi: G\rightarrow G'$ be a group homomorphism, and let $H=Ker(\phi)$.Let $a\in G$.Then the set

(1)

\begin{align} \phi^{-1}[\{\phi(a)\}]=\{x\in G|\phi(x)=\phi(a)\} \end{align}

is the left coset $aH$ of $H$,and is also the right coset $Ha$ of $H$.

So,

(2)

\begin{align} \phi^{-1}[\{\phi(11)\}]=11Ker \phi \end{align}

(3)

\begin{align} \phi^{-1}[\{\phi(11)\}]=11\{1,9, 17,33\} \end{align}

(4)

\begin{align} \phi^{-1}[\{\phi(11)\}]=\{11,19, 27,3\} \end{align}