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## Homework Statement

Consider the subset H = {0,3} of the group Z

_{6}.

(i) Prove that H is a normal subgroup of Z

_{6}.

(ii) Prove that Z

_{6}/H is isomorphic to Z

_{3}. (Give an explicit isomorphism)

## Homework Equations

In order to be a subgroup H must be closed under +

_{6}, have an identity, and have an inverse for each element.

In order to be a normal subgroup, the left cosets must equal the right cosets of H in Z

_{6}.

## The Attempt at a Solution

(i)

__H is closed under +__

_{6}I wrote out the group table:

0 +

_{6}0 = 0

3 +

_{6}0 = 3

0 +

_{6}3 = 3

3 +

_{6}3 = 0

Therefore, H is closed.

__the identity, e, is also in H__

e = 0

0 +

_{6}0 = 0

3 +

_{6}0 = 3

Therefore, e is in H.

__the inverse of a, a__

^{-1}, is in H0 +

_{6}0 = 0

3 +

_{6}3 = 0

Thus, a = a

^{-1}

Therefore, a

^{-1}is in H.

Since all three conditions hold, H must be a subgroup of Z

_{6}.

Is H a

**normal**subgroup of Z

_{6}?

left cosets:

aH =

0 +

_{6}H = {0,3}

1 +

_{6}H = {1,4}

2 +

_{6}H = {2,5)

right cosets:

Ha =

H +

_{6}0 = {0,3}

H +

_{6}1 = {1,4}

H +

_{6}2 = {2,5}

The left cosets = the right cosets of H in Z

_{6}.

Thus, H is a normal subgroup of Z

_{6}.

Is the above work correct?

Part (ii) I do not know where to even begin.

What does Z

_{6}/H mean? Is it possible to draw a group table for it?

I would really appreciate some heavy guidance on this portion.

Thank you so much for any help you can provide me!