- #1

quantumdude

Staff Emeritus

Science Advisor

Gold Member

- 5,566

- 22

Well, in 5 years of PF'ing and watching over this forum, I am finally posting my first homework question. :tongue2: I'm taking a graduate course in Algebra, and it's been 11 years since I took the undergraduate version. So, I'm going back and doing all the homework exercises in my undergrad book. I'm stuck on this one.

[itex]S_A[/itex] is the group of all permutations of a set [itex]A[/itex] under permutation multiplication. [itex]B[/itex] is a subset of [itex]A[/itex], and [itex]b[/itex] is a particular element of [itex]B[/itex]. Determine whether the given set is sure to be a subgroup of [itex]S_A[/itex] under the induced operation. Here [itex]\sigma

## Homework Statement

[itex]S_A[/itex] is the group of all permutations of a set [itex]A[/itex] under permutation multiplication. [itex]B[/itex] is a subset of [itex]A[/itex], and [itex]b[/itex] is a particular element of [itex]B[/itex]. Determine whether the given set is sure to be a subgroup of [itex]S_A[/itex] under the induced operation. Here [itex]\sigma

**=\{\sigma(x)|x \in B\}[/itex]**

And the subsets are...

[itex]H=\{\sigma\in S_A|\sigmaAnd the subsets are...

[itex]H=\{\sigma\in S_A|\sigma

**\subseteq B\}[/itex]**

[itex]K=\{\sigma\in S_A|\sigma[itex]K=\{\sigma\in S_A|\sigma

**=B\}[/itex]**

Not applicable.

First let's consider [itex]H[/itex]. The elements of [itex]H[/itex] are all of the permutations that send the elements of [itex]B[/itex] to a subset of [itex]B[/itex]. To try to grasp this, I considered an example.

Let [itex]A=\{1,2,3,4,5\}[/itex] and [itex]B=\{1,2,3\}[/itex]. Then choose a permutation [itex]\sigma_1[/itex] that satisfies the condition of membership in [itex]H[/itex].

[tex]\sigma_1=\left(\begin{array}{ccccc}1 & 2 & 3 & 4 & 5\\3 & 1 & 2 & 5 & 4 \end{array}\right)[/tex]

When I look at this, I can't see how [itex]H[/itex] could be anything other than [itex]K[/itex] itself. If the image of [itex]B[/itex] under [itex]\sigma_1[/itex] is anything other than [itex]B[/itex], then it contains elements of [itex]A[/itex] that do

The answer in the back of the book says that [itex]H[/itex] is not a subgroup of [itex]S_A[/itex], as it is not closed under taking of inverses. I do not see how that could possibly be right.

I'll leave [itex]K[/itex] alone until I get [itex]H[/itex] sorted out.

Thanks,## Homework Equations

Not applicable.

## The Attempt at a Solution

First let's consider [itex]H[/itex]. The elements of [itex]H[/itex] are all of the permutations that send the elements of [itex]B[/itex] to a subset of [itex]B[/itex]. To try to grasp this, I considered an example.

Let [itex]A=\{1,2,3,4,5\}[/itex] and [itex]B=\{1,2,3\}[/itex]. Then choose a permutation [itex]\sigma_1[/itex] that satisfies the condition of membership in [itex]H[/itex].

[tex]\sigma_1=\left(\begin{array}{ccccc}1 & 2 & 3 & 4 & 5\\3 & 1 & 2 & 5 & 4 \end{array}\right)[/tex]

When I look at this, I can't see how [itex]H[/itex] could be anything other than [itex]K[/itex] itself. If the image of [itex]B[/itex] under [itex]\sigma_1[/itex] is anything other than [itex]B[/itex], then it contains elements of [itex]A[/itex] that do

**not**belong to [itex]B[/itex]. Hence, the image would not be a subset of [itex]B[/itex].The answer in the back of the book says that [itex]H[/itex] is not a subgroup of [itex]S_A[/itex], as it is not closed under taking of inverses. I do not see how that could possibly be right.

I'll leave [itex]K[/itex] alone until I get [itex]H[/itex] sorted out.

Thanks,

Last edited: