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

[Guest2]

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 $H(x_0) \cap H(x_1)$?

 

[Euge]

Hi Guest,

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, $H(x_0)\cap H(x_1) = \{\phi\in A(S) : x_0\phi = x_0, x_1 \phi = x_1\}$.

 

[Guest2]

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.

 

[Deveno]

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).

 

[Guest2]

Thanks, http://mathhelpboards.com/members/deveno/.

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.

 

[Euge]

Hi Guest,

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 $g^{-1}\in H\cap K$.

 

[Guest2]

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.

 

[Euge]

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 $A(S)$.