Verifying Subgroup Notation of $H(x_0)$ in $A(S)

In summary, verifying subgroup notation of $H(x_0)$ in $A(S)$ involves checking if the elements in the subset $H(x_0)$ of the group $A(S)$ satisfy the necessary conditions to be considered a subgroup. This includes verifying closure under the group operation, existence of an identity element, and existence of inverses for each element. Additionally, the notation should also indicate the group operation and the identity element. If all these conditions are met, then $H(x_0)$ can be confirmed as a subgroup of $A(S)$.
  • #1
Guest2
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Let $S$ be any set, $A(S)$ the set of one-to-one mappings of $S$ onto itself, made into a group under the composition of mappings. If $x_0 \in S$, what is meant by $H(x_0) = \left\{\phi \in A(S): x_0 \phi = x_0\right\}$? The set that contains the element $\phi$ in $A(S)$ that maps $x_0$ onto itself?

How does one verify that something like this is a subgroup of $A(S)$?

For any $\phi, \phi' \in H(x_0)$, we have $x_0\phi \phi' = x_0 \phi' = x_0 $. Thus $H(x_0)$ is closed under multiplication. How do I show that for any $\phi \in H(x_0)$ we have $\phi^{-1} \in H(x_0)$?

Also if for $x_1 \ne x_0 \in H(x_1)$, we similarly define $H(x_1)$, what's $H(x_0) \cap H(x_1)$?
 
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  • #2
Hi Guest,

Guest said:
If $x_0 \in S$, what is meant by $H(x_0) = \left\{\phi \in A(S): x_0 \phi = x_0\right\}$? The set that contains the element $\phi$ in $A(S)$ that maps $x_0$ onto itself?

No, $H(x_0)$ is the set of all mappings $\phi\in A(S)$ which satisfy $x_0\phi = x_0$.

How does one verify that something like this is a subgroup of $A(S)$?

For any $\phi, \phi' \in H(x_0)$, we have $x_0\phi \phi' = x_0 \phi' = x_0 $. Thus $H(x_0)$ is closed under multiplication. How do I show that for any $\phi \in H(x_0)$ we have $\phi^{-1} \in H(x_0)$?

To show that $H(x_0)$ is a subgroup, it suffices to prove that $H(x_0)$ is nonempty, closed under the group operation, and closed under inverses. You have already shown that $H(x_0)$ is closed under composition. The set is nonempty because the identity mapping belongs to $H(x_0)$. To show closure under inverses, take $\phi\in H(x_0)$ so that $x_0\phi = x_0$. Then $x_0\phi^{-1} = (x_0\phi)\phi^{-1} = x_0(\phi\phi^{-1}) = x_0i = x_0$, showing that $\phi^{-1}\in H(x_0)$.

Also if for $x_1 \ne x_0 \in H(x_1)$, we similarly define $H(x_1)$, what's $H(x_0) \cap H(x_1)$?

If $A$ and $B$ are subsets of a set $X$, then $A\cap B$ is the set of all elements $x\in X$ such that $x\in A$ and $x\in B$. Now $\phi\in H(x_0)$ if and only if $x_0 \phi = x_0$, and $\phi\in H(x_1)$ if and only if $x_1\phi = x_1$. Therefore, $\phi\in H(x_0)\cap H(x_1)$ if and only if $x_0 \phi = x_0$ and $x_1 \phi = x_1$. Hence, $H(x_0)\cap H(x_1) = \{\phi\in A(S) : x_0\phi = x_0, x_1 \phi = x_1\}$.
 
  • #3
Hi http://mathhelpboards.com/members/euge/,

Thank you!

And $K = H(x_0)\cap H(x_1)$ is a subgroup of $A(S)$, right? Let $\psi, \phi \in K$. Then by definition, $\psi, \phi \in H(x_0)$ and $\psi, \phi \in H(x_1).$ Checking whether it's closed under the group operation, we have $x_0 \psi \phi = x_0 \phi = x_0 \in H(x_0) \subset K .$ Therefore $\psi \phi \in K$. Similarly, we have $x_1 \psi \phi$ $ = x_1 \phi = x_1 \in H(x_1) \subset K$. And again, $\psi \phi \in K$. Now, $K$ is nonempty since the identity map $i \in K$. To show show closure under inverses, it suffices that $x_0\phi^{-1} = (x_0\phi)\phi^{-1} = x_0(\phi\phi^{-1}) = x_0i = x_0$ and $x_1\phi^{-1} = (x_1\phi)\phi^{-1} = x_1(\phi\phi^{-1}) = x_1i = x_1$, thus $\phi^{-1} \in K$.

Also, is $L = H(x_0)\cup H(x_1) =\{\phi\in A(S) : x_0\phi = x_0 ~\text{or}~ x_1 \phi = x_1\}$ a subgroup of $A(S)$? I understand that when both $x_0\phi = x_0$ and $x_1 \phi = x_1$ are satisfied, we have $L = K$, so it's a subgroup of $A(S)$. But I can't wrap head around it when one of them isn't satisfied.
 
  • #4
Guest said:
Hi http://mathhelpboards.com/members/euge/,

Thank you!

And $K = H(x_0)\cap H(x_1)$ is a subgroup of $A(S)$, right? Let $\psi, \phi \in K$. Then by definition, $\psi, \phi \in H(x_0)$ and $\psi, \phi \in H(x_1).$ Checking whether it's closed under the group operation, we have $x_0 \psi \phi = x_0 \phi = x_0 \in H(x_0) \subset K .$ Therefore $\psi \phi \in K$. Similarly, we have $x_1 \psi \phi$ $ = x_1 \phi = x_1 \in H(x_1) \subset K$. And again, $\psi \phi \in K$. Now, $K$ is nonempty since the identity map $i \in K$. To show show closure under inverses, it suffices that $x_0\phi^{-1} = (x_0\phi)\phi^{-1} = x_0(\phi\phi^{-1}) = x_0i = x_0$ and $x_1\phi^{-1} = (x_1\phi)\phi^{-1} = x_1(\phi\phi^{-1}) = x_1i = x_1$, thus $\phi^{-1} \in K$.

Also, is $L = H(x_0)\cup H(x_1) =\{\phi\in A(S) : x_0\phi = x_0 ~\text{or}~ x_1 \phi = x_1\}$ a subgroup of $A(S)$? I understand that when both $x_0\phi = x_0$ and $x_1 \phi = x_1$ are satisfied, we have $L = K$, so it's a subgroup of $A(S)$. But I can't wrap head around it when one of them isn't satisfied.

The situation is analogous to that of subspaces of vector spaces (this should not be all *that* surprising: vector spaces are "groups with extra structure"):

1. For subgroups $H,K$ of a group $G$, the set $H \cap K$ is also a subgroup.

2. $H \cup K$ is rarely a subgroup, unless one subgroup contains the other. In this case, the smallest subgroup containing $H,K$ is $\langle H,K\rangle = \langle (H\cup K)\rangle$, the group *generated* by $H$ and $K$. This group is not always so *easy* to describe explictly, but basically contains all possible products of elements of the form $hk$ with $h \in H,k \in K$, so a typical element looks like:

$h_1k_1\cdots h_nk_n$, which may, or may not, simplify (this depends on how $H$ and $K$ interact, and some of the elements might be the identity).
 
  • #5
Thanks, http://mathhelpboards.com/members/deveno/.
Deveno said:
1. For subgroups $H,K$ of a group $G$, the set $H \cap K$ is also a subgroup.
I've given ago proving this. Let $a,b \in H \cap K$. Then by definition $a,b \in H$ and $a,b \in K$. Since $H$ and $K$ are subgroups, they're closed under the group operation, therefore we have $ab \in H$ and $ab \in K$, and consequently $ab \in H \cap K$. Thus $H \cap K$ is closed under the group operation. It's also nonempty since $e \in H$ and $e \in K$ therefore $e \in H \cap K$. To prove closure under inverses, we already know that $e = g g^{-1}\in H \cap K$, which by closure under group operation gives $g,g^{-1} \in H \cap K$. Thus $H \cup K$ is also closed under inverses. Thus $H \cup K$ is subgroup since it's nonempty subset of $G$ of that's closed under the group operation and inverses.
 
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  • #6
Hi Guest,

Guest said:
Let $a,b \in H \cap K$. Then by definition $a,b \in H$ and $a,b \in K$. Since $H$ and $K$ are subgroups, they're closed under the group operation, therefore we have $ab \in H$ and $ab \in K$, and consequently $ab \in H \cap K$. Thus $H \cap K$ is closed under the group operation. It's also nonempty since $e \in H$ and $e \in K$ therefore $e \in H \cap K$.

This is correct so far.

To prove closure under inverses, we already know that $e = g g^{-1}\in H \cap K$, which by closure under group operation gives $g,g^{-1} \in H \cap K$.

This needs to be cleaned up. First of all, $g$ needs to be introduced before you write $e = gg^{-1}$. Second, having a product of elements in $H\cap K$ does not imply that the individual elements lie in $H\cap K$. For example, take $G = \Bbb Z$, $H = 2\Bbb Z$, and $K = 3\Bbb Z$. Then $H\cap K = 6\Bbb Z$. Now $3 \cdot 2 = 6\in 6\Bbb Z$, but $3\notin 6\Bbb Z$ and $2\notin 6\Bbb Z$.

Start by letting $g\in H\cap K$, and then prove $g^{-1}\in H\cap K$. Since $g\in H\cap K$, then $g\in H$ and $g\in K$. Since $H$ is a closed under inverses, $g^{-1}\in H$; as $K$ is closed under inverses, $g^{-1}\in K$. Therefore $g^{-1}\in H\cap K$.
 
  • #7
Hi http://mathhelpboards.com/members/euge/,

Thanks, that was very useful to me!

One thing I'm still confused about is how to go about proving that $H(x_0)\cup H(x_1) =\{\phi\in A(S) : x_0\phi = x_0 ~\text{or}~ x_1 \phi = x_1\}$ is not a subgroup of $A(S)$. How would go on about checking for example that this is or isn't closed under group operation? For the record, this isn't from my book. I just thought it would be a natural thing to consider after the book made me consider the intersection.
 
  • #8
It is possible for $H(x_0) \cup H(x_1)$ to be a subgroup of $A(S)$. For example, let $S = \{x_0,x_1\}$. Then $A(S) = \{\phi_0,\phi_1\}$, where $\phi_0 : \begin{cases}x_0\mapsto x_0\\x_1\mapsto x_1\end{cases}$ and $\phi_1 : \begin{cases}x_0\mapsto x_1\\x_1\mapsto x_0\end{cases}$. So $H(x_0) = \{\phi_0\} = H(x_1)$, which implies $H(x_0) \cup H(x_1) = \{\phi_0\}$, a subgroup of $A(S)$.
 

FAQ: Verifying Subgroup Notation of $H(x_0)$ in $A(S)

1. What is the purpose of verifying subgroup notation in $H(x_0)$ in $A(S)$?

The purpose of verifying subgroup notation in $H(x_0)$ in $A(S)$ is to ensure that the notation accurately represents the subgroup $H(x_0)$ in the group $A(S)$. This is important for understanding the structure and properties of the subgroup and how it relates to the larger group.

2. How is subgroup notation verified in $H(x_0)$ in $A(S)$?

Subgroup notation in $H(x_0)$ in $A(S)$ is verified by checking that the elements of $H(x_0)$ satisfy the defining properties of a subgroup. This includes closure, associativity, identity, and inverses.

3. What are the defining properties of a subgroup?

The defining properties of a subgroup include closure, associativity, identity, and inverses. Closure means that the group operation on any two elements of the subgroup results in another element that is also in the subgroup. Associativity means that the group operation is the same regardless of the order in which it is performed. Identity means that there is an element in the subgroup that acts as the identity element for the group operation. Inverses means that for every element in the subgroup, there is another element that when combined with the original element using the group operation, results in the identity element.

4. Why is it important to verify subgroup notation?

It is important to verify subgroup notation to ensure that the notation accurately represents the subgroup and its properties. This allows for better understanding of the subgroup and its relationship to the larger group. It also ensures that correct mathematical operations are performed when working with the subgroup.

5. What are some common errors that may occur in subgroup notation?

Some common errors that may occur in subgroup notation include not satisfying the defining properties of a subgroup, using incorrect symbols or notation, and not properly identifying the subgroup within the larger group. It is important to carefully check for these errors when verifying subgroup notation in order to ensure its accuracy.

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