MHB How can I find the kernel of a homomorphism of finite groups?

[onie mti]

i was given that Z₈ to Z₄ is given by
f= 0 1 2 3 4 5 6 7
0 1 2 3 0 1 2 3

where f is homomorphism. how can i find the kernal K

 

[I like Serena]

i was given that Z₈ to Z₄ is given by
f= 0 1 2 3 4 5 6 7
0 1 2 3 0 1 2 3

where f is homomorphism. how can i find the kernal K

The kernel is the set of elements that map to 0...

 

[Deveno]

It would be more compact to write:

$f(k\text{ (mod }8)) = k\text{ (mod }4)$

or even more succinctly:

$f([k]_8) = [k]_4$.

Also, usually one doesn't just "declare" a map to be a homomorphism, one has to prove it. In this example that means showing:

$f([k]_8+[m]_8) = f([k]_8) + f([m]_8)$ where the addition on the left is mod 8, and the addition on the right is mod 4.

I would start by showing:

$f([k]_8 + [m]_8) = f([k+m]_8) = [k+m]_4$.

Now clearly every $k = 8t$ (every integral multiple of 8) is also a multiple of 4: $k = 4(2t)$.

This means that $f([0]_8) = [0]_4$, so we know the kernel will at least contain $[0]_8$.

Are there integral multiples of 4 that are NOT multiples of 8?

(Hint: what are odd multiples of 4 reduced mod 8? Start with:

(2n+1)(4) = 8n + 4 =...? mod 8)

 

[onie mti]

It would be more compact to write:

$f(k\text{ (mod }8)) = k\text{ (mod }4)$

or even more succinctly:

$f([k]_8) = [k]_4$.

Also, usually one doesn't just "declare" a map to be a homomorphism, one has to prove it. In this example that means showing:

$f([k]_8+[m]_8) = f([k]_8) + f([m]_8)$ where the addition on the left is mod 8, and the addition on the right is mod 4.

I would start by showing:

$f([k]_8 + [m]_8) = f([k+m]_8) = [k+m]_4$.

Now clearly every $k = 8t$ (every integral multiple of 8) is also a multiple of 4: $k = 4(2t)$.

This means that $f([0]_8) = [0]_4$, so we know the kernel will at least contain $[0]_8$.

Are there integral multiples of 4 that are NOT multiples of 8?

(Hint: what are odd multiples of 4 reduced mod 8? Start with:

(2n+1)(4) = 8n + 4 =...? mod 8)

I do not follow from the point where you said k=8t, why are we saying that

 

[Deveno]

Presumably, we are dealing with the groups $\Bbb Z_8$ and $\Bbb Z_4$.

Do you recall how these groups are defined?

The reason I put brackets around the "$k$", is because the "1" (for example) in $\Bbb Z_8$ is not "the same 1" as we have in the integers.

I "tag" the brackets with a subscript, so we know "which group they came from".

So when I write $k$ (or some other letter) without the brackets, I mean the actual integer.

Here is how we create $\Bbb Z_8$:

we send the sets:

$\{...-24,-16,-8,0,8,16,24,...\} \to [0]_8$
$\{...-23,-15,-7,1,9,17,25,...\} \to [1]_8$
...
$\{8n+k: n \in \Bbb Z\} \to [k]_8$.

So if an integer $m$ is in the SET $[k]_8$ (these are COSETS of the multiples of 8, that is cosets of the subgroup $8\Bbb Z \subseteq \Bbb Z$), this means $m = 8n + k$, for some integer $n$.

The group $\Bbb Z_8$ is referred to variously as:

"The cyclic group of order 8"
"The integers modulo 8"
"The quotient group $\Bbb Z/8\Bbb Z$".

Do you have more questions?

 

[onie mti]

Presumably, we are dealing with the groups $\Bbb Z_8$ and $\Bbb Z_4$.

Do you recall how these groups are defined?

The reason I put brackets around the "$k$", is because the "1" (for example) in $\Bbb Z_8$ is not "the same 1" as we have in the integers.

I "tag" the brackets with a subscript, so we know "which group they came from".

So when I write $k$ (or some other letter) without the brackets, I mean the actual integer.

Here is how we create $\Bbb Z_8$:

we send the sets:

$\{...-24,-16,-8,0,8,16,24,...\} \to [0]_8$
$\{...-23,-15,-7,1,9,17,25,...\} \to [1]_8$
...
$\{8n+k: n \in \Bbb Z\} \to [k]_8$.

So if an integer $m$ is in the SET $[k]_8$ (these are COSETS of the multiples of 8, that is cosets of the subgroup $8\Bbb Z \subseteq \Bbb Z$), this means $m = 8n + k$, for some integer $n$.

The group $\Bbb Z_8$ is referred to variously as:

"The cyclic group of order 8"
"The integers modulo 8"
"The quotient group $\Bbb Z/8\Bbb Z$".

Do you have more questions?

thanks a lot.