Compute factor group $\dfrac{\mathbb{Z}_4 \times \mathbb{Z}_6}{\langle(2,3)\rangle}$ - Fraleigh p. 147 Example 15.11

Solution 1:

This answers only your first question. You should ask your three questions in three separate posts.

$\def\Z#1{\Bbb Z_{#1}}$ The “temptation” here arises from the way we understand quotient groups. Consider the simpler example of $\Bbb Z/\langle 3\rangle $. Intuitively, this is the result of taking the ordinary integers, $\Bbb Z$, and forcing $3=0$. Of course $3\ne 0$, but we can ask what would happen if we were unable to distinguish between $3$ and $0$. The answer is that we get $\Z3$, which is a group in which $3$ is equal to $0$. For example, in $\Z3$, $2+1 = 0$ and $1+1+1=0$. The construction of the quotient group as a group of cosets is a formalization of this intuitive idea.

For a different example, let $S_3$ be the set of symmetries of an equilateral triangle, $e$ be the identity symmetry, and $\rho$ be the reflection of the triangle along its vertical axis. $S_3/\langle\rho\rangle$ is the set of symmetries of the triangle if we choose to view the reflection as unimportant, or less formally, if we pretend that $\rho = e$. The remaining symmetries are now just the rotations, and indeed $S_3/\langle\rho\rangle\cong \Z3$.

In $(\Z4\times\Z6)/\langle(2,3)\rangle$ we're doing something similar: We start with $\Z4\times\Z6$ and ask what we would get if $(2,3) = (0,0)$. It's easy to make the mistake that since $\Z4/\langle2\rangle \cong \Z2$ and $\Z6/\langle3\rangle \cong \Z3$ then $(\Z4\times\Z6)/\langle(2,3)\rangle\cong \Z2\times\Z3$, but this is not the case, and this is what Fraleigh is warning you about. $\langle(2,3)\rangle$ has order 2 in $\Z4\times\Z6$, so $(\Z4\times\Z6)/\langle(2,3)\rangle$ has 12 elements, but $\Z2\times\Z3$ has only 6, so it can't be right.