# Group Homomorphism & Group Order

• rputra
In summary, the conversation discusses a problem involving a group of order 21, a homomorphism, and the lack of a normal subgroup of order 3. The solution involves using Lagrange's theorem and considering the orders of the kernel and the image of the homomorphism. By ruling out the possibility of the kernel having order 3, it can be determined that the image must have order 1, and therefore the homomorphism maps all elements of the group to 1.

#### rputra

I came across this problem in class note but I was stuck:

Assume that ##G## be a group of order 21, assume also that ##G'## is a group of order 35, and let ##\phi## be a homomorphism from ##G## to ##G.'## Assume that ##G## does not have a normal subgroup of order 3. Show that ##\phi (g) = 1## for each element ##g## in ##G##.

I know that most likely I need to use Lagrange's ##|G| = |G/H| \cdot |H|## and therefore ##|H|## divides ##|G|##, and also ##G/ker (\phi) \cong \phi (G)## and therefore ##|G/ker (\phi)| = |\phi (G)|##, but after those I am lost. A gentle and friendly explanation would therefore be very much appreciated.

Thank you very much and happy holidays to all.

rputra said:
I came across this problem in class note but I was stuck:

Assume that ##G## be a group of order 21, assume also that ##G'## is a group of order 35, and let ##\phi## be a homomorphism from ##G## to ##G.'## Assume that ##G## does not have a normal subgroup of order 3. Show that ##\phi (g) = 1## for each element ##g## in ##G##.

I know that most likely I need to use Lagrange's ##|G| = |G/H| \cdot |H|## and therefore ##|H|## divides ##|G|##, and also ##G/ker (\phi) \cong \phi (G)## and therefore ##|G/ker (\phi)| = |\phi (G)|##, but after those I am lost. A gentle and friendly explanation would therefore be very much appreciated.

Thank you very much and happy holidays to all.

You've already got everything you need. ##ker(\phi)=H## is a normal subgroup of ##G##. Therefore it has order 1,3,7 or 21 since those are the divisors of 21. ##\phi(G)## is a subgroup of ##G'## so it has order 1,5,7 or 35 since those are the divisors of 35. You are given that ##|ker(\phi)|## is not 3. Just try the other possibilities. I.e. if ##|ker(\phi)|=1## then ##|\phi(G)|=21/1=21##. Is that possible?

Thank you very much! Now I get it.

## 1. What is a group homomorphism?

A group homomorphism is a function that preserves the algebraic structure of a group, meaning that it maps the group operation of one group to the group operation of another group. In other words, if G and H are groups with operations * and · respectively, a group homomorphism f from G to H satisfies the property f(a * b) = f(a) · f(b) for all elements a and b in G.

## 2. What is a group order?

The group order, also known as the cardinality of a group, is the number of elements in the group. It is denoted by |G|, where G is the group. The group order is an important property of a group as it helps classify groups into different types and determines some of their properties.

## 3. How do you determine the order of a group?

The order of a group can be determined by counting the number of elements in the group. For example, if a group contains elements A, B, C, and D, then its order is 4. In some cases, the order of a group can also be determined by using mathematical properties or formulas specific to that group.

## 4. What is the significance of group homomorphism?

Group homomorphisms are significant because they allow us to study and understand the properties of groups by looking at their similarities and differences. They also help us establish relationships between groups and can be used to solve problems in various fields such as physics, chemistry, and computer science.

## 5. Can a group homomorphism be an isomorphism?

Yes, a group homomorphism can be an isomorphism if it is both one-to-one and onto. This means that the function is injective, meaning that different elements in the domain map to different elements in the codomain, and surjective, meaning that every element in the codomain has a corresponding element in the domain. An isomorphism preserves the group structure and is considered to be a bijective homomorphism.