Hi first time posting here, I am self studying group theory and came upon the following exercise:
Let G=(Z/612Z)* a) using the Chinese rem theorem, construct an EXPLICIT isomorphism f: (Z/4Z)*x(Z/9Z)x(Z/17Z) -> G=(Z/612Z)* b) Calculate |G| and m=max{ord(x) : x is an element of G} c) Using the first two parts, gind g element of G s.t ord(g)=m
For part a, how is the given isomorphism not explicit? What would be an explicit one? For parts b,c could someone provide a solution with explanation, or give me some hints?
Thanks
An "explicit" isomorphism means that if your are given $(a,b,c)$ you can calculate (explicitely) $f(a,b,c)$. For your information the chinese remainder theorem says that the explicit function :
$$\frac{\mathbb{Z}}{612\mathbb{Z}}\rightarrow \frac{\mathbb{Z}}{4\mathbb{Z}}\times \frac{\mathbb{Z}}{9\mathbb{Z}}\times \frac{\mathbb{Z}}{17\mathbb{Z}}$$
$$x\mapsto (x\text{ mod } 4,x\text{ mod } 9,x\text{ mod } 17)$$
is an isomorphism (because $612=4\times 9\times 17$ and $4$, $9$ and $17$ are prime to each other). Finding the explicit isomorphism $f$ boils down to give the inverse function of the function above. Since $f$ is a ring isomorphism $f(a,b,c)=af(1,0,0)+bf(0,1,0)+cf(0,0,1)$, you basically only need to compute $f(1,0,0)$, $f(0,1,0)$ and $f(0,0,1)$. For instance $f(1,0,0)$ is a number divisible by both $9$ and $17$ so $f(1,0,0)=153y$, now mod $4$ this number is $y$ so $y=1$ is okay. Hence $f(1,0,0)=153$. For $f(0,1,0)=68y$ which is $5y$ mod $9$ so that $y$ must be the inverse of $5$ modulo $9$ which happens to be $2$ so $f(0,1,0)=136$. Finally $f(0,0,1)=36y$ which is $2y$ mod $17$ so that if $y=9$ then $f(0,0,1)=324$.
Hence $f(a,b,c)=153a+136b+324c$ is the explicit form. Just restrict it to the group of multiplicative inverses to get the wanted isomorphism.
For question $b$ it is easier to look at the order of elements in the group $G'= \frac{\mathbb{Z}}{4\mathbb{Z}}^*\times \frac{\mathbb{Z}}{9\mathbb{Z}}^*\times \frac{\mathbb{Z}}{17\mathbb{Z}}^*$ (a by-product of the chinese remainder theorem is precisely that $G$ and $G'$ are isomorphic so you can answer question $b$ in $G'$ rather than in $G$, it is easier). Show that each group in the product is cyclic (by explicitely giving a generating element). Deduce from this the number $m$ you are looking for.
For question $c$ when doing question $b$ you will find the $m$ by explicitely giving an element $(a_0,b_0,c_0)$ of $ \frac{\mathbb{Z}}{4\mathbb{Z}}^*\times \frac{\mathbb{Z}}{9\mathbb{Z}}^*\times \frac{\mathbb{Z}}{17\mathbb{Z}}^*$ whose order is maximal, then use question $a$ to give the corresponding element in $\frac{\mathbb{Z}}{612\mathbb{Z}}^*$.