MHB Which of the group axioms are satisfied?

[mathmari]

Hey!

I want to check the following sets with the corresponding relations if they satisfy the axioms of groups.

• $M=\mathbb{R}\cup \{\infty\}$ with the relation $\min:M\times M\rightarrow M$. It holds that $\min (a, \infty)=\min (\infty, a)=a$ for all $a\in M$.
• $M=n\mathbb{Z}=\{n\cdot a\mid a\in \mathbb{Z}\}$ for a $n\in \mathbb{N}$ with the usual multiplication of integers as the relation.
• $M=\{w,f\}$ with the relation $A\star B:=(A\implies B)$.
• $M=\{g\in G\mid \forall h \in G: g\star h=h\star g\}$ for a group $(G, \star )$ with the relation of the group $(G, \star)$ as the relation.

The axioms that we have to check are the following, aren't they?

1. Associativity : For all $a, b, c \in G$ it holds that $a \star (b \star c) = (a \star b) \star c$.
2. There is an element $e \in G$ with the property $e \star a = a$ for all $a \in G$. ($e$ is the identity of $G$).
3. For each element $a \in G$ there is $a' \in G$, such that $a'\star a = e$. ($a'$ is the inverse to $a$).

I have done the following:

• The associativity is satisfied, isn't it? Do we jusify that as follows?
  $\min (a, \min (b, c)) = \text{ minum of } \{a, b, c\} = \min(\min (a,b), c)$
  
  The identity is $e=\infty$.
  
  In this case there is no inverse.
• The associativity is satisfied, as the multiplication of integers is associative.
  
  There is no identity and there is not an inverse for each element (for example there is no inverse for the element $n$).
• Could you give me a hint for this set?
  
• Since we have the relation of the group $(G, \star)$, all the three axioms are satisfied.
  

Is everything correct? (Wondering)

 

[I like Serena]

Hey!

I want to check the following sets with the corresponding relations if they satisfy the axioms of groups.

• $M=\mathbb{R}\cup \{\infty\}$ with the relation $\min:M\times M\rightarrow M$. It holds that $\min (a, \infty)=\min (\infty, a)=a$ for all $a\in M$.
• $M=n\mathbb{Z}=\{n\cdot a\mid a\in \mathbb{Z}\}$ for a $n\in \mathbb{N}$ with the usual multiplication of integers as the relation.
• $M=\{w,f\}$ with the relation $A\star B:=(A\implies B)$.
• $M=\{g\in G\mid \forall h \in G: g\star h=h\star g\}$ for a group $(G, \star )$ with the relation of the group $(G, \star)$ as the relation.

The axioms that we have to check are the following, aren't they?

1. Associativity : For all $a, b, c \in G$ it holds that $a \star (b \star c) = (a \star b) \star c$.
2. There is an element $e \in G$ with the property $e \star a = a$ for all $a \in G$. ($e$ is the identity of $G$).
3. For each element $a \in G$ there is $a' \in G$, such that $a'\star a = e$. ($a'$ is the inverse to $a$).

Hey mathmari!

You're missing one, which I'd like to call the 0th axiom: the set has to be closed under the operation $\star$.
Furthermore, we require $e \star a = a \star e = a$, since the operation is not necessarily commutative.
Similarly we require $a'\star a = a \star a' = e$.

I have done the following:

• The associativity is satisfied, isn't it? Do we jusify that as follows?
  $\min (a, \min (b, c)) = \text{ minum of } \{a, b, c\} = \min(\min (a,b), c)$

You're using the usual property of minimum that it is supposed to be associative.
But doesn't this minimum have a special case for $\infty$ so that associativity is not immediately obvious for the cases where $\infty$ is involved?

• The identity is $e=\infty$.
  
  In this case there is no inverse.

Correct.

• The associativity is satisfied, as the multiplication of integers is associative.
  
  There is no identity and there is not an inverse for each element (for example there is no inverse for the element $n$).

There is an identity if $n=1$. For $n>1$ there is no identity.
Either way, it's not a group because even for $n=1$ the element $0$ doesn't have an inverse.

• Could you give me a hint for this set?

I assume that $w$ stands for $\text{Wahr}$? Can we make it $\{t,f\}$ then?
Then it's closed and associative with respect to the implication operator.
But it's not associative.
Can we find an identity $e$ such that for each $A$ we have that $(A\Rightarrow e) = (e\Rightarrow A) = A$?
When is an implication true exactly? (Wondering)

• Since we have the relation of the group $(G, \star)$, all the three axioms are satisfied.

Not so fast.
For starters there is an additional condition, which I like to call to 0th axiom: the set has to be closed under the operation. Is it satisfied?
We can skip associativity since that is indeed implied from $G$.
Is the identity of $G$ also contained in $M$? If so then it will also be an identity in $M$.
Does each element in $M$ have an inverse that is also in $M$?

For the record, since $M\subseteq G$ we have that $M$ is a group iff it is a subgroup of $G$.
To verify if $M$ is a subgroup it suffices to verify if it is:

• Non-empty.
• Closed under the operation $\star$.
• Closed under inverses.

(Nerd)

 

[mathmari]

You're using the usual property of minimum that it is supposed to be associative.
But doesn't this minimum have a special case for $\infty$ so that associativity is not immediately obvious for the cases where $\infty$ is involved?

Ok. And what do we do if no element is the minimum? Or do we consider in that case the usual property? (Wondering)

I assume that $w$ stands for $\text{Wahr}$? Can we make it $\{t,f\}$ then?
Then it's closed and associative with respect to the implication operator.
Can we find an identity $e$ such that for each $A$ we have that $(A\Rightarrow e) = (e\Rightarrow A) = A$?
When is an implication true exactly? (Wondering)

The identity must be T, or not? Because it holds that $(T\Rightarrow T)=T$ and $(T\rightarrow F)=F$. Is this correct? (Wondering)

About the inverse... There is no such an element, is there? (Wondering)

For starters there is an additional condition, which I like to call to 0th axiom: the set has to be closed under the operation. Is it satisfied?

Do we not get that again that $(G, \star)$ is a group? (Wondering)

Is the identity of $G$ also contained in $M$?

Let $e$ be the identity of $G$. Then we have that $e\star h=h=h\star e$, or not? From that it follows that $e$ is also contained in $M$, right? (Wondering)

Does each element in $M$ have an inverse that is also in $M$?

Let $g\in M$. Let $g^{-1}$ be the inverse of $g$. Then we have that $g^{-1}\star g=e=g\star g^{-1}$, so $g^{-1}\in M$, right? (Wondering)

 

[I like Serena]

Ok. And what do we do if no element is the minimum? Or do we consider in that case the usual property? (Wondering)

If no element is the minimum, then we do not have closure, so that M would not be a group.
However, let $a=\infty$. Then what is $\min(a,\infty)$?

The identity must be T, or not? Because it holds that $(T\Rightarrow T)=T$ and $(T\rightarrow F)=F$. Is this correct? (Wondering)

That shows that T is a left-identity. Is T also a right-identity?

Do we not get that again that $(G, \star)$ is a group? (Wondering)

Closure is indeed given for G. But shouldn't we verify it for M?

Let $e$ be the identity of $G$. Then we have that $e\star h=h=h\star e$, or not? From that it follows that $e$ is also contained in $M$, right? (Wondering)

Right!

Let $g\in M$. Let $g^{-1}$ be the inverse of $g$. Then we have that $g^{-1}\star g=e=g\star g^{-1}$, so $g^{-1}\in M$, right? (Wondering)

Shouldn't it hold for any h instead of just g? (Wondering)

 

[mathmari]

If no element is the minimum, then we do not have closure, so that M would not be a group.
However, let $a=\infty$. Then what is $\min(a,\infty)$?

If $\infty\notin \{a,b,c\}$ then we have that \begin{equation*}\min (a, \min (b,c))=\text{ Minimum of } \{a,b,c\}=\min (\min (a,b), c)\end{equation*}

If $\infty\in \{a,b,c\}$ (let $a=\infty$ then $\min (a,x)=\min (x,a)=x, \forall x$) then we have that \begin{equation*}\min (a, \min (b,c))=\min (b,c) \ \text{ and } \ \min (\min (a,b), c)=\min (b,c)\end{equation*}

Do we show in that way the associative property? (Wondering) About the inverse element: Each element $a\neq \infty$ has no inverse, since $\min (a,b)$ for each $b$ is not equal to the identity $\infty$.

Is this correct? (Wondering)

That shows that T is a left-identity. Is T also a right-identity?

Is is just the left-identity, so there is no identity, right?

 

[I like Serena]

If $\infty\notin \{a,b,c\}$ then we have that \begin{equation*}\min (a, \min (b,c))=\text{ Minimum of } \{a,b,c\}=\min (\min (a,b), c)\end{equation*}

If $\infty\in \{a,b,c\}$ (let $a=\infty$ then $\min (a,x)=\min (x,a)=x, \forall x$) then we have that \begin{equation*}\min (a, \min (b,c))=\min (b,c) \ \text{ and } \ \min (\min (a,b), c)=\min (b,c)\end{equation*}

Do we show in that way the associative property? (Wondering) About the inverse element: Each element $a\neq \infty$ has no inverse, since $\min (a,b)$ for each $b$ is not equal to the identity $\infty$.

Is this correct? (Wondering)

Is is just the left-identity, so there is no identity, right?

Yep. All correct. (Nod)

 

[Evgeny.Makarov]

I assume that $w$ stands for $\text{Wahr}$? Can we make it $\{t,f\}$ then?
Then it's closed and associative with respect to the implication operator.

I am not sure about associativity.

 

[mathmari]

I am not sure about associativity.

The associativity does not hold: $f\star (t\star f)=f\star f=t$ but $(f\star t)\star f=t\star f=f$, right?

($a\star b$ is defined as $a\implies b$)

 

[I like Serena]

• $M=\{g\in G\mid \forall h \in G: g\star h=h\star g\}$ for a group $(G, \star )$ with the relation of the group $(G, \star)$ as the relation.

To finish it up.

As you already said, we have $e\in M$ since $e\star h = h\star e = h$.
It also means that $M$ is non-empty.

If $g_1,g_2\in M$, then:
$\forall h \in G: (g_1 \star g_2)\star h=g_1\star (g_2\star h)=g_1\star (h\star g_2)=(g_1\star h)\star g_2=(h\star g_1)\star g_2 = h\star (g_1\star g_2)$
So $g_1\star g_2 \in M$, and therefore $M$ is closed under the operation $\star$.

If $g\in M$, then:
$\forall h \in G: g\star h=h\star g \quad\Rightarrow\quad
g^{-1}\star(g\star h)\star g^{-1} = g^{-1}\star(h\star g)\star g^{-1} \quad\Rightarrow\quad
h\star g^{-1}=g^{-1}\star h$
So $g^{-1}\in M$, and therefore $M$ is closed under inverses.

Thus $M$ is a subgroup of $G$ and is therefore a group on its own. (Thinking)

 

[mathmari]

Thanks a lot! (Smile)