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!!!
Using surjectivity, there exists a such that where has order 8.
Therefore, , which means that .
Now, . Remembering that is a finite group, we have that for some the minimal such number ( has some finite order, which is this number ).
Clearly, this means that . By the minimality of , we conclude that 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, , 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).
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 .
So, . This element has an order, and it is finite because is finite. Call the order of `n'. That is, and there does not exist with .
Now, , so has order n, and so has order . That is, . Thus, has order 8.