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Consider $f: \mathbb Z_8 \rightarrow \mathbb Z_4$ given by

$$\begin{pmatrix} 0&1&2&3&4&5&6&7 \\ 0&1&2&3&0&1&2&3 \end{pmatrix}$$

Verify the $f$ is a homomorphism, find its kernal $K$, and list the cosets of $K$.

I have verified that this is a homomorphism by a Cayley Table, and the kernal $K=\{0,4\}$.

Now it says list the cosets of $K$. I am not sure of this.

Is the coset of $K$ the elements that $f$ does not map into the identity, i.e.: $\{1,2,3,5,6,7\}$?

Or, since the kernal of $f$ is a normal subgroup of the domain, in this case $\mathbb Z_8$ (i.e.: $K$ is subgroup of $\mathbb Z_8$), are the cosets of the kernal equal to $Ka$, where $a$ is any element of $\mathbb Z_8$?

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    $\begingroup$ Note the spelling "Kernel" not "Kernal". $\endgroup$ – Mark Bennet Nov 7 '14 at 20:57
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Since $|ker|=2$, so there will be $4$ elements in the coset. Now follow the usual rule for forming cosets of a subgroup, i.e., $aH=\{ah\mid a\in G\}$. Thus for this case the coset corresponding to the element $0,4$ is $K$ itself. For example, for $1$ its $1+K=\{1,5\}$. In a similar way cosets for other elements of $\mathbb{Z}_8$ can be formed, keeping in mind the operation on $\mathbb{Z}_8$. See that the cosets corresponding to $1$ and $5$ and $3$ and $7$ are the same.

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Hint: Note that the permutation corresponds to the map

$f:a\to a(\mod 4)$

Clearly, Kernel of the map is elements of $\mathbb{Z}_8$ which map to $0$ which are $0$ and $4$, which you have already written. $K=\{ 0,4\}$

What are cosets? Suppose $N$ is a normal subgroup of $G$

Then $G/N=\{ a\oplus_8N : a\in G\}$. Note $\oplus_8$ is addition modulo 8.

The distinct cosets are

$\mathbb{Z}_8/K=\{ K,1\oplus_8K,2\oplus_8K,3\oplus_8K\}$

The number of distinct cosets is given by $|\mathbb{Z}_8|/|K|$

Note

$K=\{ 0,4\}$

$1\oplus_8K=\{ 1,5\}$

$2\oplus_8K=\{ 2,6\}$

$3\oplus_8K=\{ 3,7\}$

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  • $\begingroup$ That doesn't help me find the cosets of the kernal. I am confused as to the what is a coset of the kernal. I have only seen cosets of a subgroup of a group. $\endgroup$ – Al Jebr Nov 7 '14 at 20:23
  • $\begingroup$ I'll edit accordingly. $\endgroup$ – Swapnil Tripathi Nov 7 '14 at 20:24
  • $\begingroup$ @JohannFranklin The Kernel of a homomorphism is a subgroup - so there is no new kind of coset here. $\endgroup$ – Mark Bennet Nov 7 '14 at 20:56

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