How would I show that there are no onto homomorphisms from Q8 to Z4? I have to show that there are no onto homomorphisms from _ to _, including Q8 to Z4. 
How would I show that? Also can someone explain what onto means again, my teacher didn't explain it well to me. 
 A: We can put
$$Q_8=\{1,-1,i,j,k,-i,-j,-k\}\;\;,\;\;\Bbb Z_4:=\{0,1,2,3\}$$
Suppose we have a group epimorphism 
$$f:Q_8\to \Bbb Z_4\Longrightarrow Q_8/\ker f\cong\Bbb Z_4$$
But then $\,|\ker f|=2\,$ and the only subgroup of order $\,2\,$ in $\,Q_8\,$ is its center $\,\{1,-1\}\,$ , so 
$\,Q_8/\{1,-1\}\cong\Bbb Z_4\,$. But this is impossible since for any $\,q\in Q_8\setminus\{1,-1\}\,$ we have that 
$\,q^2=-1\in\{1,-1\} \Longrightarrow Q_8/\{1,-1\}\,$ is a $\,2-\,$elementary group (=a group where every element 
squared equals the unit) and, thus, it cannot be cyclic of order four.
Added: We can shorten the above noting that since  $\{1,-1\}=Z(Q_8)\,$ , and since no group modulo its center can be cyclic non-trivial, then we're done.
A: Basically, you do it by contradiction.  So suppose $f:Q_8\to Z_4$ is an onto homomorphism, and pick $a\in Q_8$ where $f(a)=1$.  Then $f(a)+f(a)=2\not=0$, but $f(a)+f(a)=f(a*a)$ since it's a homomorphism, and $a*a=1$ for any $a\in Q_8$.  We know that $f(1)=0$ for any homomorphism, so $2=f(1)=0$, and we have a contradiction.
