To which group H is the Quotient Group $\frac{GL_n(Z_p) }{SL_n(Z_p)}$ isomorphic? So I need to find a surjective homomorphism  $\phi:  GL_n(\Bbb Z_p) \to H $ s.t. $ker\phi = SL_n(\Bbb Z_p)$ and use the First Isomorphism Theorem to find $\phi' :\frac{GL_n(Z_p) }{SL_n(Z_p)} \to H $ I'm not sure how to find $\phi$
 A: If we have a homomorphism $f:GL_n(K) \to H$, it has the property that
$$f(AB)=f(A)f(B).$$
Additionally, we have that $f(A)=e \in H$ if and only if $A \in SL_n(K)$.
So pick an arbitrary matrix $X$ with determinant $\det(X)$ and consider the pair of matrices $A = \frac{X}{\det(X)^{1/n}},$ $B= \det (X)^{1/n} I$. (Assume for now that an $n$th root exists. If we have $GL_3(R)$ this is the case, so we'll pretend we are dealing with that general linear group instead.) Notice that $A \in SL_n(K)$. 
We then have
$$f(X)=f(AB)=f(A)f(B)= f(\det (X)^{1/n} I)$$
because $f(A)=e$.
That is, $f(X)=g(\det(X))$ must be some function of the determinant of $X$, which itself is a homomorphism of the underlying field $K$. That is, $f: GL_n(K) \to K \to H$.
We might as well try the simplest case $g:K \to K$ be the identity function. Here, it is very easy to show this defines a homomorphism (regardless of $K$) whose kernel is identically the special linear group.
A: Define a mapping f:Gln(Zp) - U(n)(i.e Z(p)/{0}) such that f(A) = det(A) for all A in Gln(Zp) then it is an onto homomorphism with kernel = SLn(Zp).(Kernel={A in Gln(zp) : f(A)(det(A)) = identity of U(n) (which is 1)}= Sln(Zp) . Now apply first isomorphism theorem - the quptient group GLn(Zp)/SLn(Zp) is isomorphic to U(p).
Order(Gln(Zp)/Sln(Zp))=O(U(n))=p-1 if p is prime .
