Group theory conjectures

Question

After completing the following problem, ''if $\phi$ is a homomorphism from $\mathbb{Z}_{30}$ onto a group of order $5$, determine $\ker(\phi)$," I have made two conjectures I would like some feedback on. If you are very confident in group theory please feel free to skip to the bulleted conjectures at the end.

Recall the definition of the kernel of a group homomorphism $\phi$: the kernel of $\phi$ is $ker(\phi)=\{g\in G \text{ : } \phi(g)=e_H\}$ where $e_H$ is the identity in H.

So to determine the $ker(\phi)$, we find all elements mapped to $0$ by $\phi$.

Consider the fact that there can certainly be more than one group homomorphism $\phi$ that maps $\mathbb{Z}_{30}$ onto a group of order $5$. However, note that a group of prime order, which $5$ is, is cyclic. There exists only one cyclic group of finite order, $\mathbb{Z}_n$. Hence the group homomorphism $\phi$ maps $\mathbb{Z}_{30}$ onto $\mathbb{Z}_5$.

Recall the example in counting homomorphisms done at the end of class. We stated that homomorphisms between cyclic groups are completely determined by where it sends the generators. Note that any $m<n$ is a generator of $\mathbb{Z}_n$ if $(m,n)=1$, i.e. $m,n$ are relatively prime. Hence every element in $\mathbb{Z}_5$ is a generator for $\mathbb{Z}_5$ since $5$ is prime. Recall that if $\phi \text{ : }G\rightarrow H$ is a group homomorphism and $G,H$ are cyclic, then $\phi$ sends a generator of a $G$ to a generator of a group $H$. Since there are only $5$ possible generators for $\mathbb{Z}_5$ we check the $5$ possible homomorphisms.

Let the generator of concern for $\mathbb{Z}_{30}$ be $1_{30}$. We will be utilizing a subscript with elements to denote where the element lives, $x_{30}$ for $x\in \mathbb{Z}_{30}$ and $x_5$ for $x\in \mathbb{Z}_{5}$. The $5$ possible homomorphisms send, \begin{eqnarray*} \phi_0 \text{ : } 1_{30} &\longmapsto& 0_5 \\ \phi_1 \text{ : } 1_{30} &\longmapsto& 1_5 \\ \phi_2 \text{ : } 1_{30} &\longmapsto& 2_5 \\ \phi_3 \text{ : } 1_{30} &\longmapsto& 3_5 \\ \phi_4 \text{ : } 1_{30} &\longmapsto& 4_5, \end{eqnarray*} where the subscript on the homomorphisms indicates where $\phi(1)$ is sent.

Consider the trivial homomorphism $\phi_0$ that maps $\mathbb{Z}_{30}$ onto $\mathbb{Z}_5$,

$\phi_0 = \left( \begin{array}{cccccccccccccc} 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & \dots & 28 & 29 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \dots & 0 & 0 \\ \end{array} \right)$

For the four remaining homomorphisms, we proceed by trial and error:

Consider $\phi_1$ to be the group homomorphism that maps $\mathbb{Z}_{30}$ onto $\mathbb{Z}_5$ in the following way:

$\phi_1 = \left( \begin{array}{cccccccccccccc} 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & \dots & 28 & 29 \\ 0 & 1 & 2 & 3 & 4 & 0 & 1 & 2 & 3 & 4 & 0 & \dots & 3 & 4 \\ \end{array} \right)$

We see that $ker(\phi_1)=\{0,5,10,15,20,25\}$. Also this is a homomorphism since: $\phi_1(e_G)=e_H$ where $G=\mathbb{Z}_{30}$ and $H=\mathbb{Z}_5$, and also $\phi_1(a+b)=\phi_1(a)+\phi_1(b)$ for all $a,b \in \mathbb{Z}_{30}$.

Consider $\phi_2$ to be the group homomorphism that maps $\mathbb{Z}_{30}$ onto $\mathbb{Z}_5$ in the following way:

$\phi_2 = \left( \begin{array}{cccccccccccccc} 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & \dots & 28 & 29 \\ 0 & 2 & 4 & 3 & 0 & 1 & 2 & 4 & 3 & 0 & 1 & \dots & 3 & 0 \\ \end{array} \right)$

$\phi_2$ is not a homomorphism since $\phi_2(3+4)=\phi_2(7)=4$, but $\phi_2(3)+\phi_2(4)=3$.

However, consider a different $\phi_2$. Let $\phi_2$ be the group homomorphism that maps $\mathbb{Z}_{30}$ onto $\mathbb{Z}_5$ in the following way:

$\phi_2 = \left( \begin{array}{cccccccccccccc} 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & \dots & 28 & 29 \\ 0 & 2 & 1 & 4 & 3 & 0 & 2 & 1 & 4 & 3 & 0 & \dots & 4 & 3 \\ \end{array} \right)$

We see that $ker(\phi_2)=\{0,5,10,15,20,25\}$. Also this is a homomorphism since: $\phi_2(e_G)=e_H$ where $G=\mathbb{Z}_{30}$ and $H=\mathbb{Z}_5$, and also $\phi_2(a+b)=\phi_2(a)+\phi_2(b)$ for all $a,b \in \mathbb{Z}_{30}$.

Note the difference between the two $\phi_2$ homomorphisms. Of the first five numbers, none are mapped to themselves as occurred in the first $\phi_2$. However this may not be the only requirement of the homomorphism.

Consider $\phi_2$ to be the group homomorphism that maps $\mathbb{Z}_{30}$ onto $\mathbb{Z}_5$ in the following way:

$\phi_2 = \left( \begin{array}{cccccccccccccc} 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & \dots & 28 & 29 \\ 0 & 2 & 1 & 4 & 3 & 2 & 1 & 4 & 3 & 2 & 1 & \dots & 3 & 2 \\ \end{array} \right)$

$\phi_2$ is not a homomorphism since $\phi_2(6+3)=\phi_2(9)=2$, but $\phi_2(6)+\phi_2(3)=5$.

Now note the difference between the successful $\phi_2$ and the non successful. The $\phi_2$ that is a homomorphism sends the identity to the identity, the generator to the generator, but also, each element of the codomain $\mathbb{Z}_5$ occurs the same amount of times. This occurs since the results of the homomorphism, ``cycle" in a sense, through the values of the codomain. Notice that the unsuccessful $\phi_2$ directly above does not do this. It maps the identity to the identity but then the identity never appears again. If we combine this property with the property discovered above, that the first five numbers of the domain need to not be sent to themselves, we begin to see that the kernel of these homomorphisms must always be $\{0,5,10,15,20,25\}$.

The same property comes up with a homomorphism $\phi_3$. We have,

$\phi_3 = \left( \begin{array}{cccccccccccccc} 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & \dots & 28 & 29 \\ 0 & 3 & 1 & 4 & 2 & 0 & 3 & 1 & 4 & 2 & 0 & \dots & 4 & 2 \\ \end{array} \right)$

As predicted $ker(\phi_3)=\{0,5,10,15,20,25\}$. Also this is a homomorphism since: $\phi_3(e_G)=e_H$ where $G=\mathbb{Z}_{30}$ and $H=\mathbb{Z}_5$, and also $\phi_3(a+b)=\phi_3(a)+\phi_3(b)$ for all $a,b \in \mathbb{Z}_{30}$.

The same property comes up with a homomorphism $\phi_4$. We have,

$\phi_4 = \left( \begin{array}{cccccccccccccc} 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & \dots & 28 & 29 \\ 0 & 4 & 1 & 2 & 3 & 0 & 4 & 1 & 2 & 3 & 0 & \dots & 2 & 3 \\ \end{array} \right)$

As predicted $ker(\phi_4)=\{0,5,10,15,20,25\}$. Also this is a homomorphism since: $\phi_4(e_G)=e_H$ where $G=\mathbb{Z}_{30}$ and $H=\mathbb{Z}_5$, and also $\phi_4(a+b)=\phi_4(a)+\phi_4(b)$ for all $a,b \in \mathbb{Z}_{30}$.

Hence, barring the the trivial homomorphism $\phi_0$, we see that the kernel of all other homomorphisms is $\{0,5,10,15,20,25\}$. Note that with is isomorphic to $\mathbb{Z}_6$. We know that the kernel of a group is a subgroup, and this checks out

Based on the above exploration, we make the following claims:

• Let $f$ be an onto homomorphism between two finite cyclic groups, $f \text{ : } \mathbb{Z}_n \rightarrow \mathbb{Z}_m$, where $m<n$ and $(m,n)\neq 1$. Then we have that $ker(f)=\{0+mx\}$ for some $x\in \mathbb{Z}$ and $mx<n$.

• The number of homomorphisms between two finite cyclic groups, $\mathbb{Z}_n \rightarrow \mathbb{Z}_m$, where $m<n$ and $(m,n)\neq 1$, is equal to $(m,n)$.

Answer 1

You keep talking about "homomorphisms" which turn out not to be homomorphisms. Your first $\phi_2$ has $\phi_2(1)=2$, $\phi_2(2)=4$ and $\phi_2(3)=3$. If $\phi_2$ were a homomorphism, then we would have $$\phi_2(2)=\phi_2(1+1)=\phi_2(1)+\phi_2(1)=2+2=4,$$ (good), $$\phi_2(3)=\phi_2(2+1)=\phi_2(2)+\phi_2(1)=4+2=1$$ (as we are in $\Bbb Z_5$), which doesn't agree with your original $\phi_2$, so that isn't a homomorphism. But it shows us how to create one; next, $$\phi_2(4)=\phi_2(3+1)=\phi_2(3)+\phi_2(1)=1+2=3,$$ $$\phi_2(5)=\phi_2(4+1)=\phi_2(4)+\phi_2(1)=3+2=0,$$ etc.

The principle is that when we know what $\phi(1)$ is, we can determine what the homomorphism $\phi$ is. So, if $\phi(1)=4$, then $\phi(2)=4+4=3$. So your $\phi_4$ is not a homomorphism.

Anyway, for each homomorphism $\phi$, $\phi(5)=\phi(1)+\cdots+\phi(1)$ (five summands) which equals zero. Then $\phi(10)=\phi(5)+\phi(5)=0+0=0$ etc. The kernel contains $\{0,5,10,\ldots,25\}$. But the kernel of a surjective homomorphism has order $30/5$, so that must actually be the kernel.

Answer 2

These are true and pretty straightforward. I didn't read through all the examples because I think you're making this too complicated for yourself. In the first bullet, note that the existence of the onto homomorphism forces $m|n$. Since an onto homomorphism must take a cyclic generator (which you can take to be $1$) to a cyclic generator, precisely the $m^{th}$ multiples of the generator will be in the kernel.

For the second item, note that the homomorphism can only take a cyclic generator to the subgroup of order $(m,n)$ in the target. It can take it to any element of that subgroup, and there are $(m,n)$ ways to do this.

The main properties used here are the a homomorphism from a cyclic group to any other group is completely determined by the target of any generator, and the only restriction on the target of that generator is that its order must divide the order of the generator (and this restriction only applies if the source cyclic group is finite).