Answer to Question #107070 in Abstract Algebra for Garima Ahlawat

Question

Give a non trivial element of the ring Z10 / <4bar> , with justification.

Answer 1

The given ring is ,

$$\frac{\Z_{10}}{<\bar4>}=\{ \bar b +<\bar4>:\bar b\in \Z_{10} \}=G(say)$$

Where $<\bar4>= \{ \bar0,\bar2,\bar4,\bar6,\bar8\}$ .

Then a element $x=\bar b+<\bar4>\in G$ is called non trivial if $x\neq<\bar4>$ .

Again ,we known that $\bar a+<\bar4>=<\bar4> \iff \bar a\in <\bar4>$

as a coset .

We can easily prove it ,

Suppose that $\bar a+<\bar4>=<\bar4>$

Then ,$\bar a+\bar0 \in \bar a +<\bar4> =<\bar4>$ .

Conversely ,assume that $\bar a\in <\bar4>$ .

Since ,$<\bar4>$ is a subgroup ,therefore by closer property

$\bar a+\bar y\in <\bar4> \forall \ \bar y \in <\bar4> \implies \bar y +<\bar4>\subseteq<\bar4>$ .

Let $\bar y\in <\bar4>.$ Since $\bar a ,\bar y \in <\bar 4>$ therefore $\bar y-\bar a\in<\bar4>.$

Thus , $\bar y = \bar0+\bar y=(\bar a -\bar a)+\bar y=\bar a +(\bar y-\bar a)\in \bar a +<\bar4 >$

Hence,$\bar a +<\bar 4>=<\bar 4>$ .

Since , $\bar3 \notin <\bar4> \implies \bar3+<\bar4>$ is a non trivial element of G.