1. ## Homomorphism question

Let PHI: G -> K be a homomorphism and G finite. Show that if K has an element of order 8 then G has an element of order 8.

Attempt:

I am not really sure how to proceed since PHI is only a homomorphism I do not know if there is an element in G such that it maps to the element in K with order 8.

You guys have any hints?

EDIT::
PHI is also ONTO!!!

2. Originally Posted by mulaosmanovicben
Let PHI: G -> K be a homomorphism and G finite. Show that if K has an element of order 8 then G has an element of order 8.

Attempt:

I am not really sure how to proceed since PHI is only a homomorphism I do not know if there is an element in G such that it maps to the element in K with order 8.

You guys have any hints?

As you wrote it the claim is false: just let $\displaystyle K=C_8=$ the cyclic group of order 8,

$\displaystyle G=S_3\,,\,\phi:G\rightarrow K\,,\,\,\phi(g):= 1\,\,\forall g\in G$

Tonio

3. yes sorry you are quite right!! I forgot to mention that it is also surjective (onto)

so far I have y in K, |y| = 8 and PHI(x)=y, since G is finite |y| must divide |x| so 8 divides |x| or |x| = 8k

now what?

4. Originally Posted by mulaosmanovicben
yes sorry you are quite right!! I forgot to mention that it is also surjective (onto)

so far I have y in K, |y| = 8 and PHI(x)=y, since G is finite |y| must divide |x| so 8 divides |x| or |x| = 8k

now what?
Using surjectivity, there exists a $\displaystyle g \in G$ such that $\displaystyle g\phi = k$ where $\displaystyle k \in K$ has order 8.

Therefore, $\displaystyle (g^8) \phi = 1$, which means that $\displaystyle g^8 \in ker(\phi)$.

Now, $\displaystyle ker(\phi) \leq G$. Remembering that $\displaystyle ker(\phi)$ is a finite group, we have that $\displaystyle (g^8)^n = 1$ for some $\displaystyle n$ the minimal such number ($\displaystyle g^8$ has some finite order, which is this number $\displaystyle n$).

Clearly, this means that $\displaystyle (g^n)^8=1$. By the minimality of $\displaystyle n$, we conclude that $\displaystyle g^n$ has order 8.

Note that this does not work for infinite groups - you need the kernel to be finite get get that element of finite order. For example, $\displaystyle F_2$, the free group on 2 generators, surjects onto every 2-generated group (by definition) but is torsion free (it contains no no-trivial elements of finite order).

5. does this work as well:

G/KerPHI is isomorphic to K with isomorphism g*kerPHI --> PHI(g). Then consider y in K where |y|=8. y=PHI(x). so x*kerPHI --> y so x*kerPHI has same order as y so x*kerPHI has order 8 which means x has order 8.

that seems too easy?

Also I do not think i fully understand your proof. Why does g^n have order 8? Why could it not be lower?

6. Originally Posted by mulaosmanovicben
does this work as well:

G/KerPHI is isomorphic to K with isomorphism g*kerPHI --> PHI(g). Then consider y in K where |y|=8. y=PHI(x). so x*kerPHI --> y so x*kerPHI has same order as y so x*kerPHI has order 8 which means x has order 8.

that seems too easy?

Also I do not think i fully understand your proof. Why does g^n have order 8? Why could it not be lower?
so x*kerPHI has order 8 which means x has order 8' - this is not true. What is true is that $\displaystyle (xker(\phi))^8 = 1ker(\phi) \Rightarrow x^8 \in Ker(\phi)$.

So, $\displaystyle h:=x^8 \in Ker(\phi)$. This element has an order, and it is finite because $\displaystyle ker(\phi)<G$ is finite. Call the order of $\displaystyle h$ n'. That is, $\displaystyle h^n=1$ and there does not exist $\displaystyle m<n$ with $\displaystyle h^m=1$.

Now, $\displaystyle h=x^8$, so $\displaystyle x^8$ has order n, and so $\displaystyle x$ has order $\displaystyle 8n$. That is, $\displaystyle x^{8n}=1$. Thus, $\displaystyle x^n$ has order 8.

7. why does order of x is 8n imply order of x^n is 8? Why couldnt it be smaller

8. nevermind i understand now! If order of x^n was anything smaller than 8 it would contradict minimality of n just like you said! thanks!!!