# Proving Existence of Element of Order n in Finite Group via Homomorphism

• strangequark
In summary, the conversation discusses a homomorphism from a finite group G onto G', and how to prove that G contains an element of order n if G' has an element of order n. The conversation suggests looking at properties of preimages and subgroup divisibility theorem to find a solution. An analogy is given of a large gear driving a small gear with every tooth of the large gear striking the same tooth of the small gear on every revolution.
strangequark

## Homework Statement

Suppose that $$\phi$$ is a homomorphism from a finite group G onto G' and that G' has an element (g') of order n. Prove that G has an element of order n.

## Homework Equations

for a homomorphism,
1) $$\phi(a*b)=\phi(a)*\phi(b)$$
2) $$\phi(a^{n})=(\phi(a))^{n}$$
3) $$\phi(e_{G})=e_{G'}$$

## The Attempt at a Solution

It is clear to me that G will contain some non-identity element, say g, which is the preimage of g'. By property 2) that I listed above, $$g^{8}$$ is obviously an element of the kernal of G, and the homomorphism is not the trivial map because $$g^{n}$$ for 0<n<8 is not the identity in G and doesn't map to the identity in G'. Basically, I'm seeing that $$g^{8}$$ maps to the identity in G', but I don't understand why this implies that $$g^{8}=e$$...

I would really appreciate a kick in the right direction... thanks

You're completely right. It doesn't imply g^8 = e. You're going to have to do a bit more work than just finding an element of the preimage of g'. Unfortunately, I'm having trouble coming up with the exact proof off the top of my head, so I can't give you much of a kick in the right direction. Look at some properties of preimages of subgroups maybe?

Let T(a)^8=I', then a has to have an order n*8, and (a^n)^8=I
This is also summarized by some subgroup divisibility theorem; but it's been to long for me to remember.
For an Image: Imagine a large gear driving a small gear but every tooth of the large gear strikes the same tooth of the small gear on every revolution.

## 1. What is a group homomorphism?

A group homomorphism is a function that maps elements from one group to another while preserving the group structure. In other words, the operation of the first group is preserved in the second group.

## 2. How is a group homomorphism different from a group isomorphism?

While a group homomorphism preserves the group structure, a group isomorphism is a bijective homomorphism, meaning it is both one-to-one and onto. This means that a group isomorphism not only preserves the group structure, but also preserves the individual elements and their relationships between the two groups.

## 3. What is the kernel of a group homomorphism?

The kernel of a group homomorphism is the set of elements in the domain group that are mapped to the identity element in the codomain group. It is denoted as Ker(f) and is a subgroup of the domain group.

## 4. How do you determine if a function is a group homomorphism?

In order for a function to be a group homomorphism, it must satisfy the property f(ab) = f(a)f(b) for all elements a and b in the domain group. This means that the function must preserve the group operation.

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

Yes, a group homomorphism can also be an isomorphism if it is both one-to-one and onto. This means that it not only preserves the group structure, but also preserves the individual elements and their relationships between the two groups.

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