Consider the homomorphism $\phi:(\mathbb{Z}/27\mathbb{Z})\times(\mathbb{Z}/9\mathbb{Z})\to\mathbb{Z}/3\mathbb{Z}$ defined by $(a,b)\mapsto a+b$. What is the isomorphism class of $\ker\phi$?
The elements of $\ker\phi$ consist of ordered pairs $(a,b)$ where $a+b=3n$ for a positive integer $n\leq9$. There are $27\times 3=9\times 9=81$ elements of $\ker\phi$, since given the first coordinate, there are 3 options for the second coordinate (e.g., choosing 1 for the first coordinate of an element of $\ker\phi$, we can have 2, 5, or 8 in the second coordinate). However, I'm not sure how to find the group structure on this set, except to say that $\ker\phi$ is a normal subgroup of $(\mathbb{Z}/27\mathbb{Z})\times(\mathbb{Z}/9\mathbb{Z})$. (But in this case we have normality anyway because the groups are abelian.)
A hint would be very appreciated.
Best Answer
Let $u,v$ be the classes of $(1,-1)$ and $(0,3)$ in $\Bbb Z/27\Bbb Z\times\Bbb Z/9\Bbb Z$. Their orders are respectively $27$ and $3$, and every element of $\ker\phi$ writes uniquely $$au+cv,\quad 0\le a<27,\quad0\le c<3,$$ hence $$\ker\phi=\langle u\rangle\oplus\langle v\rangle\simeq\Bbb Z/27\Bbb Z\times\Bbb Z/3\Bbb Z.$$ Edit: proof (upon request by two comments) that every element in $\ker\phi$ writes uniquely $$au+cv,\quad 0\le a<27,\quad0\le c<3.$$ Let $x\in[0,27),y\in[0,9)$ be integers such that $3\mid x+y$. Then, the class of $(x,y)$ in $\Bbb Z/27\Bbb Z\times\Bbb Z/9\Bbb Z$ writes as above iff $$a\in[0,27),\quad c\in[0,3),\quad x\equiv a\bmod{27},\text{ and }y\equiv-a+3c\bmod9,$$ i.e. iff $$a=x,\quad c\in[0,3),\text{ and }x+y\equiv3c\bmod9,$$ i.e. $a=x$ and $c$ is the unique integer in $[0,3)$ such that $c\equiv\frac{x+y}3\bmod3$.