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$Z$ is the set of integers.i.e $\mathbb Z=\{.....,-2,-1,0,1,2,....\}$ . We know, $(\mathbb Z,+)$ satisfies the group axioms.Hence it is a group as well. $(4\mathbb Z,+)$ is a subgroup of $\mathbb Z$ where $4\mathbb Z=\{....,-8,-4,0,4,8,....\}$. Some of its cosets are $4\mathbb Z+0,4\mathbb Z+1,4\mathbb Z+2$ and $ 4\mathbb Z+3$ .

Someone said me that if we construct a set of this cosets then the set satisfies group axioms.Hence, $Z/4\mathbb Z$ is a quotient group.He gave me some information for the validity of his statement.His given information is in the following.

Closure: ($4\mathbb Z+2)+(4\mathbb Z+3)=4\mathbb Z+1$.

Associativity: $[(4\mathbb Z+2)+(4\mathbb Z+3)]+(4\mathbb Z+1)=4\mathbb Z+2$.

$(4\mathbb Z+2)+[(4\mathbb Z+3)]+(4\mathbb Z+1)]=4\mathbb Z+2$.

Identity element: $(4\mathbb Z+3)+4\mathbb Z=4\mathbb Z+3$

Inverse element: $(4\mathbb Z+3)+(4\mathbb Z+1)=4\mathbb Z$

I have understood his statement but I have a little bit confusion. My confusion is: is there any inverse element of $(4\mathbb Z+2)$ in $Z/4\mathbb Z$? If so, what is that?

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  • $\begingroup$ $2+4\mathbb Z$ is its own inverse. $\endgroup$ – drhab Jun 6 '15 at 13:41
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In general, in a group (as opposed to in a ring), the inverse of a coset is the coset of the inverse. In your case, the inverse of $4 \Bbb Z + 2$ would be $4 \Bbb Z -2$, but a moment of reflection will show that this is in fact the same as $4 \Bbb Z + 2$.

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$4 \mathbb Z + 2$ is the inverse of itself.

More generally, if $G$ is any group and $H$ is normal in $G$. Then you defined an induced operation $$\begin{align}\bullet: \frac{G}{H} \times \frac{G}{H} &\to \frac{G}{H}\\(gH, kH) &\mapsto g\bullet k\,H\end{align}$$

As $H \triangleleft G$ this function is well-defined. With this in mind we have that the set of cosets is indeed a group where the idendity element of $\frac{G}{H}$ is $H$, and the inverse element is $g^{-1}H$.

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