Is $|a_1\rangle\langle a_2|\otimes |b_1\rangle\langle b_2| = |a_1b_1\rangle \langle a_2b_2|$? Why or why not? I'll be using the Dirac bra-ket notation. Is $|a_1\rangle\langle a_2|\otimes |b_1\rangle\langle b_2|$ the same as $(|a_1\rangle\otimes|b_1\rangle)(\langle a_2|\otimes \langle b_2|)$? I'm not sure which property of tensor products applies here. $|a_1\rangle,|a_2\rangle,|b_1\rangle,|b_2\rangle$ are all (say) coordinate vectors in $\Bbb C^N$. $|a_1\rangle\langle a_2|$ stands for the outer product between $|a_1\rangle$ and $\langle a_2|$ and similarly $|b_1\rangle\langle b_2|$.
 A: As usual, we represent ket $\newcommand{\ket}[1]{\lvert{#1}\rangle}\newcommand{\bra}[1]{\langle{#1}\rvert} \ket{a_1}$ by a column vector $\mathbb{C}^N=\mathbb{C}^{N\times 1}$ and bra $\bra{a_2}$ by a row vector $\mathbb{C}^N=\mathbb{C}^{1\times N}$.  The outer product of two kets $\ket{a_1}$ and $\ket{a_2}$ is the ket-bra matrix product $\ket{a_1}\bra{a_2}\in\mathbb{C}^{N\times N}$, and similarly for $b_1,b_2$.
Then $\ket{a_1}\bra{a_2}\otimes\ket{b_1}\bra{b_2}=(\ket{a_1}\otimes\ket{b_1})(\bra{a_2}\otimes\bra{b_2})$ is just an instance of the mixed product property of the Kronecker product for matrices.
If we do not assume finite-dimensional state-space, of course the way to do this is to bracket this operator with states and by appealing to definition
$$
\newcommand{\braket}[2]{\langle {#1}\mid{#2}\rangle}
\bra{\phi_1\psi_1}(\ket{a_1b_1}\otimes\bra{a_2b_2})\ket{\phi_2\psi_2} = \braket{\phi_1\psi_1}{a_1b_1}\braket{a_2b_2}{\phi_2\psi_2}=\braket{\phi_1}{a_1}\braket{\psi_1}{b_1}\braket{a_2}{\phi_2}\braket{b_2}{\psi_2}
$$
and similarly
$$
\bra{\phi_1\psi_1}(\ket{a_1}\bra{a_2}\otimes\ket{b_1}\bra{b_2})\ket{\phi_2\psi_2}=\braket{\phi_1}{a_1}\braket{a_2}{\phi_2}\cdot\braket{\psi_1}{b_1}\braket{b_2}{\psi_2}
$$
so they are equal for all $\bra{\phi_1},\bra{\psi_1},\ket{\phi_1},\ket{\psi_1}$, so equal.
A: Yes, it's the same, in a sense. The property you seek is the existence of a canonical isomorphism between spaces $L(X_1,X_2)\otimes L(Y_1,Y_2)$ and $L(X_1\otimes Y_1,X_2\otimes Y_2)$, for $X_1,X_2,Y_1,Y_2$ being arbitrary vector spaces. This isomorphism  $$ \iota : L(X_1,X_2)\otimes L(Y_1,Y_2) \rightarrow L(X_1\otimes Y_1,X_2\otimes Y_2)$$
can be defined by the property $$ \big(\iota(f\otimes g)\big)(x \otimes y) = f(x) \otimes g(y) $$
In your case, you have $f =|a_1\rangle\langle a_2|$, $g= |b_1\rangle\langle b_2|$. You can check that for $x=|\phi\rangle$, $y=|\psi\rangle$ :
\begin{align} \big(\iota (|a_1\rangle\langle a_2|\otimes |b_1\rangle\langle b_2|)\big)|\phi\otimes\psi\rangle &= |a_1\rangle\langle a_2|\phi\rangle \otimes |b_1\rangle\langle b_2|\psi\rangle = \\
&= |a_1\rangle \otimes |b_1\rangle \langle a_2|\phi\rangle\langle b_2|\psi\rangle = \\
&=|a_1\otimes b_1\rangle \langle a_2\otimes b_2|\phi\otimes\psi\rangle \end{align}
so 
$$ \iota (|a_1\rangle\langle a_2|\otimes |b_1\rangle\langle b_2|) = |a_1\otimes b_1\rangle \langle a_2\otimes b_2|$$
