Group mono-, endo-, iso-, homomorphism

[mathmari]

Hey!

Let $(G, \#), \ (H, \square )$ be groups. Show:

1. For $(g,h), (g',h')\in G\times H$ we define the operation $\star$ on $G\star H$ as follows:
   \begin{equation*}\star: (G\times H)\times (G\times H,\star), \ \left ((g,h), (g',h')\right )\mapsto (g\# g', h\square h')\end{equation*} Then $(G\times H, \star)$ is a group.
2. The map $G\rightarrow G\times H, \ g\mapsto (g, e_H)$ is a monomorphism, where $e_H$ is the neutral element in $H$.
3. The map $G\times H\rightarrow G, \ (g,h)\mapsto g$ is an endomorphism.
4. Let $G=\mathbb{Z}/4\mathbb{Z}$ and $H=\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$. Does it hold that $G\sim H$ ?

For 1:
We have to show that the four axioms (closure, associativity, identity, inverse) are satisfied. For 2:
Let $\phi$ be the name of that map. Then we have that $\phi (g\#g')=(g\#g', e_H)$. How could we continue? For 3:
Let $\psi$ be the name of that map. Then we have that $\psi ((g,h)\star (g',h'))=(g\#g', h\square h')$. How could we continue? For 4:
We have to define $f:G\rightarrow H$ and check if this map is bijective, right? :unsure:

 

[I like Serena]

For 1:
We have to show that the four axioms (closure, associativity, identity, inverse) are satisfied.

Hey mathmari!

Yep. (Nod)

Btw, shouldn't it be:

• For $(g,h), (g',h')\in G\times H$ we define the operation $\star$ on $G\times H$ as follows:
  \begin{equation*}\star: (G\times H)\times (G\times H)\to (G\times H), \ \left ((g,h), (g',h')\right )\mapsto (g\# g', h\square h')\end{equation*} Then $(G\times H, \star)$ is a group.

? (Wondering)

For 2:
Let $\phi$ be the name of that map. Then we have that $\phi (g\#g')=(g\#g', e_H)$. How could we continue?

Shouldn't we check if it is equal to $\phi(g)\star \phi(g')$?

For 3:
Let $\psi$ be the name of that map. Then we have that $\psi ((g,h)\star (g',h'))=(g\#g', h\square h')$. How could we continue?

Shouldn't we check if it is equal to $\psi((g,h))\star\psi((g',h'))$?

For 4:
We have to define $f:G\rightarrow H$ and check if this map is bijective, right?

Shouldn't we check if the map is a homomorphism as well?

 

[mathmari]

Btw, shouldn't it be:

• For $(g,h), (g',h')\in G\times H$ we define the operation $\star$ on $G\times H$ as follows:
  \begin{equation*}\star: (G\times H)\times (G\times H)\to (G\times H), \ \left ((g,h), (g',h')\right )\mapsto (g\# g', h\square h')\end{equation*} Then $(G\times H, \star)$ is a group.

? (Wondering)

Ahh yes!

It is closed since the result of $(g,h)\star (g',h')$ with $(g,h), (g',h')\in G\times H$ is again in $G\times H$.

We have that \begin{align*}&[(g,h)\star (g',h')]\star (\tilde{g},\tilde{h})=(g\# g', h\square h')\star (\tilde{g},\tilde{h})=(g\# g'\#\tilde{g}, h\square h'\square \tilde{h}) \\ &(g,h)\star [(g',h')\star (\tilde{g},\tilde{h})]=(g,h)\star (g'\# \tilde{g},h'\square \tilde{h})= (g\#g'\# \tilde{g},h\square h'\square \tilde{h})\end{align*}
So the associativity follows.

The identity is $(e_G,e_H)$ since $(g,h)\star (e_G,e_H)=(g\# e_G, h\square e_H)=(g,h)$ and $ (e_G,e_H)\star (g,h)=(e_G\# g, e_H\square h)=(g,h)$.

Since $G$ and $H$ are groups, they have closed by inverses. So $g^{-1}\in G$ and $h^{-1}\in H$ the inverses of $g$ and $h$ respectively. Then the inverse of $(g,h)\in G\times H$ is $(g^{-1}, h^{-1})$ since $(g,h)\star (g^{-1}, h^{-1})=(g\# g^{-1}, h\square h^{-1})=(e_G,e_H)$ and $ (g^{-1},h^{-1})\star (g,h)=(g^{-1}\# g, h^{-1}\square h)=(e_G,e_H)$.

So it follows that $(G\times H, \star)$ is a group.

Shouldn't we check if it is equal to $\phi(g)\star \phi(g')$?

We have that $\phi(g)\star \phi(g')=(g,e_H)\star (g',e_H)=(g\#g', e_H\square e_H)$. It holds that $e_H\square e_H=e_H$ because $e_H$ is the neutral element of $H$, right?
Since $\phi (g\#g')=(g\#g', e_H)$ and so $\phi(g)\star \phi(g')=\phi (g\#g')$

Further we have to show that $\phi$ is injective. Let $\phi(g)= \phi(g')$. Then we get $(g,e_H)=(g',e_H)$ and it follows that $g=g'$.
Therefore $\phi$ is injective and so it is a monmorphism.

Shouldn't we check if it is equal to $\psi((g,h))\star\psi((g',h'))$?

We have that $\psi((g,h))\star\psi((g',h'))=g\star g'$. Is this correct? Isn't the operation $\star$ defined at elements of $G\times H$ but we have elements of $G$. :unsure:

Shouldn't we check if the map is a homomorphism as well?

Ok.. But how is the map defined? :unsure:

 

[I like Serena]

We have that $\psi((g,h))\star\psi((g',h'))=g\star g'$. Is this correct? Isn't the operation $\star$ defined at elements of $G\times H$ but we have elements of $G$.

Ah, I intended $\psi((g,h))\#\psi((g',h'))$. (Blush)

Ok.. But how is the map defined?

I think we have to come up with a map ourselves, don't we?
Or otherwise prove that there is no such map.
We might check what the order of the elements are to see if there is such a map.

 

[mathmari]

Ah, I intended $\psi((g,h))\#\psi((g',h'))$. (Blush)

Ahh! So we have that $\psi((g,h))\#\psi((g',h'))=g\#g$. We also have that $\psi ((g,h)\star (g',h'))=\psi((g\#g', h\square h'))=g\#g'$.
Therefore $\psi$ is an homomorphism.

It is left to show that $\psi$ is surjective. For each $g\in G$ there is an element $(g,h)\in G\times H$ for some $h\in H$ such that $\psi ((g,h))=g$.
Therefore $\psi$ is an endomorphism.

Is that correct? :unsure:

I think we have to come up with a map ourselves, don't we?
Or otherwise prove that there is no such map.
We might check what the order of the elements are to see if there is such a map.

We have that the orders of elements of $\mathbb{Z}_2 \times \mathbb{Z}_2$ are $|(0, 0)| = 1, |(0, 1)| = 2, |(1, 0)| = 2, |(1, 1)| = 2$, while the orders of the elements of $\mathbb{Z}_4$ are $|0| = 1, |1| = 4, |2| = 2$, and $|3| = 4$. Since at $\mathbb{Z}_4$ there is an element of order 4 but not in $\mathbb{Z}_2 \times \mathbb{Z}_2$, these cannot be isomorphic, right? :unsure:

 

[I like Serena]

Therefore $\psi$ is an endomorphism.

Is that correct?

It's not an endomorphism is it?
An endomorphism is a homomorphism from a group to the same group. (Worried)

these cannot be isomorphic, right?

Indeed. :)

 

[mathmari]

It's not an endomorphism is it?
An endomorphism is a homomorphism from a group to the same group. (Worried)

Oh I meant epimorphism, it was typo...Sorry!

Indeed. :)

In this case, for these $G$ and $H$ do we consider the operations as above or the addition? :unsure:

 

[I like Serena]

Oh I meant epimorphism, it was typo...Sorry!

Then it is correct. :)

In this case, for these $G$ and $H$ do we consider the operations as above or the addition?

For $G$ it is indeed addition.
For $H$ it would be the $\star$ as above where each individual operation is addition, wouldn't it?

 

[mathmari]

For $G$ it is indeed addition.
For $H$ it would be the $\star$ as above where each individual operation is addition, wouldn't it?

So in this case we consider $\#$ to be $+$.
But at $H$ above we have $\square$, why do we consider here $\star$ ? I got stuck right now. Or do you mean at $G\star H$ to consider $\star$ ? :unsure:

 

[I like Serena]

So in this case we consider $\#$ to be $+$.
But at $H$ above we have $\square$, why do we consider here $\star$ ? I got stuck right now. Or do you mean at $G\star H$ to consider $\star$ ?

The symbols $G$ and $H$ are confusing here since we had the group $(G\times H,\star)$ above, which is not what we have in this question 5.
Let's rename the groups in question 5 to $\tilde G = (\mathbb Z/4\mathbb Z,+)$ and $\tilde H=(\mathbb Z/2\mathbb Z\times \mathbb Z/2\mathbb Z,\star)$, where $(a,b)\star(c,d)=(a+b,c+d)$.
So the operation for $\tilde G$ is simply $+$, which is the default operation on $\mathbb Z/n\mathbb Z$. No need to involve $\#$ here.
And both $\#$ and $\square$ are $+$ for $\tilde H$. Instead of $\star$, we might also simply use $+$.

Does it make sense now? (Wondering)