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I am having trouble finding an isomorphism between $U(15)$ and $\mathbb Z_4\oplus \mathbb Z_2$. What we learned in class to create an isomorphism we typically find the generators and go from there. However, I can't find a generator for wither group. I am supposed to list where each element of $U(15)$ is mapped and also check three cases such that $\varphi(ab) = \varphi(a)\varphi(b)$. I know the elements in $$U(15)=\{1,2,4,7,8,11,13,14\}$$ and in $$\mathbb Z_4\oplus \mathbb Z_2=\{(0,0),(0,1),(1,0),(1,1),(2,0),(2,1),(3,0),(3,1)\}.$$ I found the order of $(0,0)=1$,$(0,1)=2$, $(1,0)=4$, $(1,1)=4$,$(2,0)=2$,$(2,1)=2$,$(3,0)=4$,$(3,1)=4$. In $U(15)$ $4,11,14$ have order $2$ and $2,7,8,13$ have order $4$ and $1$ has order $1$.

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  • $\begingroup$ Please format your post with proper MathJax markup. $\endgroup$ – anomaly Mar 1 '17 at 1:33
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You have that $\mathbb Z_4\oplus\mathbb Z_2$ is generated by two commuting elements, one of order $4$ and one of order $2$ (namely, $(1,0)$ and $(0,1)$).

On $U(15)$, we want generators $a$ of order $4$ and $b$ of order $2$; if we take $a=2$, the table below shows that $b\ne4$; taking $b=11$, $$ \begin{matrix} a=2\\ a^2=4\\ a^3=8\\ b=11\\ ab=7\\ a^2b=14\\ a^3b=13 \end{matrix} $$ So $a,b$ generate $U(15)$ and we can define $$ \phi(n,m)=2^n11^m. $$

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