What is the isomorphism class of the kernel of a homomorphism of cyclic groups?

Question

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.

Answer 1

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 $c\equiv\frac{x+y}3\bmod3$.

Answer 2

Lets call G the group. A good way of narrowing down is using a consecuense of the first theorem of isomorphism as the cardinalities are finite and the morphism is surjective, we have:

$$|\ker(\phi)|\cdot |img(\phi)| = |G|$$

Where we get $|\ker(\phi)|= 3^4$

Then if we take $a$ and $b$ module 3, you can see that if $a\equiv 0 \implies b\equiv 0$ as $a+b\equiv 0$ analogously $a\equiv 1 \implies b\equiv 2$ and $a\equiv 2 \implies b\equiv 1$ and this cover all cases and more over the cardinalities argument will prove that this are all cases.

The preivuos analisis gives the elements in the kernel $$\{(3k,3l),(3k+1,3l+2),(3k+2,3l+1)|k,l \in \mathbb{Z}\}$$

And for each form is a simple count that are $3^3$ elements, and three (disjoint) forms, having $3^4$ elements ie all the elements in the kernel.

Finally by the way we find the elements in the kernel is isomorphic to:

$\mathbb{Z}/27\mathbb{Z}\times\mathbb{Z}/3\mathbb{Z}$.

That if beacuse of each $a \in \mathbb{Z}$ there are three elements in the kernel $(a,-a),(a,3-a),(a,6-a)$.