42 Groups Acting on Themselves by Left Multiplication - Cayley's Theorem

--23 ✅ 2025-02-22

2

Question

List the elements of $S_{3}$ as $1, (1 2), (2 3), (1 3), (1 2 3), (1 3 2)$ and label these with the integers $1, 2, 3, 4, 5, 6$ respectively. Exhibit the image of each element of $S_{3}$ under the Left Regular Representation of $S_{3}$ into $S_{6}$ .

Proof

Here:

Label	Element in $S_{3}$
1	1
2	(1 2)
3	(2 3)
4	(1 3)
5	(1 2 3)
6	(1 3 2)
Then applying $σ : S_{3} \to S_{6}$ is done by considering $σ_{σ \in S_{3}}$ applied to each $σ \in S_{3}$ in our set.

Clearly $σ_{1} (σ) = σ$ . For example $σ_{1} (1) = 1, σ_{1} (2) = 2, \dots$ . What we can do is use the labels and see the group action on each element. Namely, let $x_{1} = 1, x_{2} = (1 2), \dots$ . Then we get the following table:

v First (arg) \| Second ( $σ$ )>	$x_{1}$	$x_{2}$	$x_{3}$	$x_{4}$	$x_{5}$	$x_{6}$	Associated $σ \in S_{6}$
$x_{1} = 1$	1	2	3	4	5	6	$1$
$x_{2} = (1 2)$	2	1	5	6	3	4	$(1 2) (3 5) (4 6)$
$x_{3} = (2 3)$	3	6	1	5	4	2	$(1 3) (2 6) (4 5)$
$x_{4} = (1 3)$	4	5	6	1	2	3	$(1 4) (2 5) (3 6)$
$x_{5} = (1 2 3)$	5	...
$x_{6} = (1 3 2)$	6	...
... and while we could fill out the whole table, really we're looking for each $σ \in S_{6}$ associated with each element $σ_{i}$ via this group action. Notice that because $σ$ is a homomorphism them for example:

\begin{aligned} σ_{(1 2 3)} & = σ_{(1 2) (2 3)} \\ = σ_{(1 2)} σ_{(2 3)} \\ = (1 2) (3 5) (4 6) \cdot (1 3) (2 6) (4 5) \\ = (1 5 6) (2 4 3) \end{aligned}

\begin{aligned} σ_{(1 3 2)} & = σ_{(1 2) (1 3)} \\ = (1 2) (3 5) (4 6) \cdot (1 4) (2 5) (3 6) \\ = (1 6 5) (2 3 4) \end{aligned}

Thus we can find what we were looking for:

Element in $S_{3}$	Image of that element within $S_{6}$
$1$	$σ_{1} = 1$
$(1 2)$	$σ_{(1 2)} = (1 2) (3 5) (4 6)$
$(1 3)$	$σ_{(1 3)} = (1 4) (2 5) (3 6)$
$(2 3)$	$σ_{(2 3)} = (1 3) (2 6) (4 5)$
$(1 2 3)$	$σ_{(1 2 3)} = (1 5 6) (2 4 3)$
$(1 3 2)$	$σ_{(1 3 2)} = (1 6 5) (2 3 4)$

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9

Question

Prove that if $p$ is a prime and $G$ is a Group of Order (Groups) $p^{α}$ for some $α \in Z^{+}$ , then every Subgroup of Index (Groups) $p$ is a Normal Subgroup in $G$ . Deduce that every group of order $p^{2}$ has a normal subgroup of order $p$ .

Proof

Suppose $| G | = p^{α}$ where $p$ is prime and $α \in Z^{+}$ . Consider some subgroup $H \leq G$ where $| G : H | = p$ . Then by Lagrange's Divisibility Theorem of Order of Subgroups we have:

\frac{| G |}{| H |} = p = \frac{p^{α}}{| H |} \Rightarrow | H | = p^{α - 1}

Thus, this implies we do an induction. If $α = 1$ then $| H | = 1$ which is the trivial subgroup (which itself is normal). If $α \in Z^{+}$ satisfies the proposition then for some $H \leq G$ where $| G | = p^{α + 1}$ and $| G : H | = p$ (still) then $| H | = p^{α}$ which itself must be a normal subgroup using the inductive hypothesis (apply $H$ to be the subgroup of itself. Then $H$ must be normal by the hypothesis, as desired). This completes the induction.

For the deduction, let $| G | = p^{2}$ . Then we just need to construct a subgroup $H \leq G$ such that $| H | = p$ , and then we can apply our proposition with $α = 2$ . Namely, to construct this, since $G$ is a group then it is non-empty. Furthermore, it must have a non-identity element (since $p \geq 2$ ). Then $\exists x \in G$ such that $x \neq 1$ . Then choose $H = ⟨ x ⟩$ . Notice that since $x \in H$ then $| H | > 1$ . Furthermore, by Finite Groups, Order of x divides Order of G, then $| H |$ divides $G$ , thus $| H | = p$ as desired.

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10

Question

Prove that every non-Abelian Group of Order (Groups) 6 has a non-Normal Subgroup of order 3. Use this to classify groups of order 6. [Hint: Produce an injective Homomorphism (or Group Morphism) into $S_{3}$ ].

Proof

Let $G$ be non-abelian where $| G | = 6$ . Notice first that since $G$ is non-abelian, then $\exists g_{1}, g_{2} \in G (g_{1} g_{2} \neq g_{2} g_{1})$ . Consider making the subgroup $⟨ g_{1} ⟩$ . Notice that by Finite Groups, Order of x divides Order of G, then $| ⟨ g_{1} ⟩ |$ must divide 6. Clearly the order cannot be 1 (since $g_{1}$ is a non-identity element and is in $⟨ g_{1} ⟩$ ) but it also cannot be 6 (since if it was then since $g_{2} \in G$ then there is some power $n$ where $g_{2} = g_{1}^{n}$ , but that would contradict $g_{1}, g_{2}$ not commuting as $g_{1} g_{2} = g_{1} g_{1}^{n} = g_{1}^{n} g_{1} = g_{2} g_{1}$ ). By a similar argument, we've shown that $g_{2} \notin ⟨ g_{1} ⟩$ , so then another subgroup can be made $⟨ g_{2} ⟩$ .

Therefore, either $| g_{1} | = 2$ or $| g_{1} | = 3$ and the same is for $g_{2}$ . One could easily find contradiction in $| G |$ to show that one has order 2 and the other had order 3 (if not, then that implies $| G | \geq 9$ or $| G | \leq 4$ as contradictions). Without loss of generality, suppose $| g_{1} | = 3$ . That implies $| G : ⟨ g_{1} ⟩ | = 2$ . By Paul's Theorem then $\frac{G}{⟨ g_{1} ⟩} ⊴ G$ . But by the orders equaling $⟨ g_{2} ⟩$ then $⟨ g_{2} ⟩ ⊴ G$ consequently (by being an Isomorphism, namely $\frac{G}{⟨ g_{1} ⟩} ≅ ⟨ g_{2} ⟩$ ). I'm doing this to show that while there will be one subgroup that follows to be normal, the other one won't which we can always choose!

Similarly, notice that $⟨ g_{1} ⟩$ is non-normal because $g_{2} ⟨ g_{1} ⟩ \neq ⟨ g_{1} ⟩ g_{2}$ due to the lack of commutativity. Thus choosing $⟨ g_{1} ⟩$ as our subgroup will always work! If the other case where $| g_{1} | = 2$ occurred, then $⟨ g_{1} ⟩ ⊴ G$ but we could choose instead $⟨ g_{2} ⟩$ as our non-normal subgroup.

Lastly, let's make an injective homomorphism into $S_{3}$ via Cayley's Theorem $σ : G \to S_{3}$ . Namely, without loss of generality that $| g_{1} | = 3$ and $| g_{2} | = 2$ as described before. We note some equivalences:

$g_{1}^{3} = 1$
$g_{2}^{2} = 2$
Thus notice that this is similar to $D_{6} = ⟨ r, s | r^{3} = s^{2} = 1, s r = r^{2} s ⟩$ . If we could show that $r \equiv g_{1}$ and $s \equiv g_{2}$ then notice that the first relation is correct. We'd want to show that $g_{2} g_{1} = g_{1}^{2} g_{2}$ . Notice we can eliminate the candidates for $g_{2} g_{1} = x_{5}$ :
1. It can't be 1 since if $g_{2} g_{1} = 1$ then $g_{2} = g_{1}^{- 1}$ which do commute.
2. It can't be $g_{2}$ because by Cancellation Theorem then $g_{1} = 1$ which is incorrect.
3. If can't be $g_{1}$ since then right-cancellation gives $g_{2} = 1$ which is incorrect.
4. If $g_{2} g_{1} = g_{1}^{2} \Rightarrow g_{2} = g_{1}$ which is wrong as they don't commute.
5. It can't be $g_{1} g_{2}$ as then $g_{2} g_{1} = g_{1} g_{2}$ contradicts the lack of commutativity.
  Thus $g_{2} g_{1} = g_{1}^{2} g_{2}$ and thus we can treat $G$ as just $D_{6}$ ! This helps us fill the table as now $g_{2} g_{1} = g_{1}^{2} g_{2}$ is an equivalence we can use. Some other equivalences that are derived then:
$g_{2} g_{1} g_{2} \equiv s r s = s s r^{2} = r^{2} \equiv g_{1}^{2}$
$g_{1} g_{2} g_{1} \equiv r s r = r r^{2} s = r^{3} s = s \equiv g_{2}$
Thus $G ≅ D_{6}$ , but one can show that $D_{6} ≅ S_{3}$ via the map $φ : D_{6} \to S_{3}$ defined by:

$d \in D_{6}$	$\mapsto p \in S_{3}$
$1$	$1$
$r$	$(1 2 3)$
$s$	$(1 2)$
Since these are the same generators and have the same equivalence relations. Notice $(1 2 3)^{3} = 1 = (1 2)^{2}$ but for the last relation:

\begin{aligned} r^{2} s = s r & \equiv (1 2 3)^{2} (1 2) = (1 2) (1 2 3) \\ \equiv (1 3 2) (1 2) = (2 3) \\ \equiv (2 3) = (2 3) \end{aligned}

Thus this is a valid isomorphism!

This suggests that all:

Abelian Groups of order 6 are congruent to $\frac{Z}{6 Z}$ because there is either elements of order 2, 3, or 6 (from our argumentation before), but by our same reasoning one would show that $(g_{1} g_{2})^{6} = 1$ and thus $| g_{1} g_{2} | = 6$ which implies $G$ is Cyclic Group and thus congruent.
Non-abelian groups are congruent to $S_{3}$ by our reasoning above.
classifying all groups of order 6!

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