MHB Does \( A_4 \) Have a Subgroup of Order 6?

[mathmari]

Hey!

I want to show that $A_4$ has no subgroup of order $6$. $A_4$ is the group of even permutations of $S_4$.

Suppose that $A_4$ has a subgroup of order $6$.

Could you give me some hints how we could get a contradiction? (Wondering)

 

[Deveno]

Hint: look at the cycle types

3-cycles
2+2-cycles
the identity.

Are any of these of order 6?

That eliminates cyclic subgroups of order 6.

So the only possibility would be a subgroup isomorphic to $S_3$, which is generated by an element of order 2, and an element of order 3.

For two such elements, say $a$ (of order 2), and $b$ (of order 3), we have to have:

$ba = ab^2$ (why?)

Can you prove this doesn't happen?

 

[mathmari]

Hint: look at the cycle types

3-cycles
2+2-cycles
the identity.

Are any of these of order 6?

The 3-cycles have order $3$.

The 2+2-cycles have order $2$.

The identity has order $4$.

Right? (Wondering)

So none of these is of order 6.

So the only possibility would be a subgroup isomorphic to $S_3$, which is generated by an element of order 2, and an element of order 3.

Why is that the only possibility when there are no cyclic subgroups of order $6$? (Wondering)

 

[Deveno]

The 3-cycles have order $3$.

The 2+2-cycles have order $2$.

The identity has order $4$.

Right? (Wondering)

So none of these is of order 6.


Why is that the only possibility when there are no cyclic subgroups of order $6$? (Wondering)

Up to isomorphisim, there are just two groups of order $6$, the cyclic group of order $6$, and the dihedral (= isomorphic to $S_3$) group of order $6$.

The latter is generated by any element of order $2$, together with any element of order $3$.

In such a group, the product of two distinct elements of order $2$, is an element of order $3$ (that is, in $S_3$ the product of any two *different* transpositions is a $3$-cycle, or in $D_3$ the product of any two *different* reflections is a non-trivial rotation).

 

[mathmari]

In such a group, the product of two distinct elements of order $2$, is an element of order $3$ (that is, in $S_3$ the product of any two *different* transpositions is a $3$-cycle, or in $D_3$ the product of any two *different* reflections is a non-trivial rotation).

I haven't really understood that... Could you explain it to me? (Wondering)

 

[Deveno]

Let us examine $S_3$, first.

Suppose we have $(a\ b)$ and $(b\ c)$ (two transpositions in $S_3$ that are distinct must share one letter in common they move, and for any transposition $(a\ b) = (b\ a)$, so we can always put the shared letter at the end of the first transposition, and the beginning of the second transposition).

Composing $(a\ b)(b\ c)$ right-to-left, we have:

$a \mapsto a \mapsto b$
$b \mapsto c \mapsto c$
$c \mapsto b \mapsto a$

so the product is $(a\ b\ c)$, a three-cycle.

In $D_3 = \langle r,s: r^3 = s^2 = 1, sr = r^2s\rangle$, we have:

$(r^ks)(r^ms) = r^k(sr^m)s = (r^k)(r^{-m}s)s = r^{k-m}s^2 = r^{k-m}$, with $k,m \in \{0,1,2\}$.

Since $k \neq m$ (we are assuming distinct reflections) $k-m \neq 0$, so $k -m = -2,-1,1$ or $2$, and since:

$r^2 = r^{-1}$, the product above is either $r$ or $r^2$, both of which have order $3$.

 

[mathmari]

Let us examine $S_3$, first.

Suppose we have $(a\ b)$ and $(b\ c)$ (two transpositions in $S_3$ that are distinct must share one letter in common they move, and for any transposition $(a\ b) = (b\ a)$, so we can always put the shared letter at the end of the first transposition, and the beginning of the second transposition).

Composing $(a\ b)(b\ c)$ right-to-left, we have:

$a \mapsto a \mapsto b$
$b \mapsto c \mapsto c$
$c \mapsto b \mapsto a$

so the product is $(a\ b\ c)$, a three-cycle.

In $D_3 = \langle r,s: r^3 = s^2 = 1, sr = r^2s\rangle$, we have:

$(r^ks)(r^ms) = r^k(sr^m)s = (r^k)(r^{-m}s)s = r^{k-m}s^2 = r^{k-m}$, with $k,m \in \{0,1,2\}$.

Since $k \neq m$ (we are assuming distinct reflections) $k-m \neq 0$, so $k -m = -2,-1,1$ or $2$, and since:

$r^2 = r^{-1}$, the product above is either $r$ or $r^2$, both of which have order $3$.

I see... (Thinking)

Up to isomorphisim, there are just two groups of order $6$, the cyclic group of order $6$, and the dihedral (= isomorphic to $S_3$) group of order $6$.

The latter is generated by any element of order $2$, together with any element of order $3$.
.

We have that $D_3$ is generated by any element of order $2$, together with any element of order $3$.
Is $D_3$ also generated by any element of order $2$, together with any element of order $3$ ? (Wondering)

 

[Deveno]

I see... (Thinking)

We have that $D_3$ is generated by any element of order $2$, together with any element of order $3$.
Is $D_3$ also generated by any element of order $2$, together with any element of order $3$ ? (Wondering)

Is there a typo in this post? You seem to have written the same thing twice (Wondering).

 

[mathmari]

Is there a typo in this post? You seem to have written the same thing twice (Wondering).

I meant:

"Is $S_3$ also generated by any element of order $2$, together with any element of order $3$ ?"

(Blush)

 

[Deveno]

I meant:

"Is $S_3$ also generated by any element of order $2$, together with any element of order $3$ ?"

(Blush)

Yes. You can prove this (although it's kind of tedious). But these are the very sorts of properties that are preserved by isomorphisms.

 

[mathmari]

Does it always stand that if we have two isomorphic groups and the one of them is generated by an element of order $n$ then the other one must also be generated by an element order $n$ ? (Wondering)

 

[Deveno]

Does it always stand that if we have two isomorphic groups and the one of them is generated by an element of order $n$ then the other one must also be generated by an element order $n$ ? (Wondering)

Think about it: if $G$ is generated by a (single) element of order $n$, then $G$ is cyclic of order $n$.

But what I think you are really asking is:

If $h: G \to G'$ is an isomorphism, and $G = \langle a,b\rangle$, do we have $G' = \langle h(a),h(b)\rangle$?

The answer is yes, AND: the orders of $h(a),h(b)$ in $G'$ are the orders of $a,b$ in $G$. Note this is only true of ISOMORPHISMS, for mere homomorphisms we cannot make such sweeping conclusions (for example, $h$ might be the homomorphism that maps everything to the identity of $G'$, which certainly does NOT preserve orders).

 

[mathmari]

Think about it: if $G$ is generated by a (single) element of order $n$, then $G$ is cyclic of order $n$.

But what I think you are really asking is:

If $h: G \to G'$ is an isomorphism, and $G = \langle a,b\rangle$, do we have $G' = \langle h(a),h(b)\rangle$?

The answer is yes, AND: the orders of $h(a),h(b)$ in $G'$ are the orders of $a,b$ in $G$. Note this is only true of ISOMORPHISMS, for mere homomorphisms we cannot make such sweeping conclusions (for example, $h$ might be the homomorphism that maps everything to the identity of $G'$, which certainly does NOT preserve orders).

Ah ok... I see.. (Thinking) I haven't really understood why it is not possible that a subgroup of $A_4$ is isomorphic to $S_3$. (Wondering)

Up to isomorphisim, there are just two groups of order $6$, the cyclic group of order $6$, and the dihedral (= isomorphic to $S_3$) group of order $6$.

The latter is generated by any element of order $2$, together with any element of order $3$.

In such a group, the product of two distinct elements of order $2$, is an element of order $3$.

The elements of $A_4$ are the identity, 3-cycles and 2-transpositions, right? (Wondering)

Do we have to prove that the product of a 3-cycle and 2-transpositions is not of order $3$ ? (Wondering)

 

[Deveno]

Ah ok... I see.. (Thinking) I haven't really understood why it is not possible that a subgroup of $A_4$ is isomorphic to $S_3$. (Wondering)



The elements of $A_4$ are the identity, 3-cycles and 2-transpositions, right? (Wondering)

Do we have to prove that the product of a 3-cycle and 2-transpositions is not of order $3$ ? (Wondering)

In both $S_3$ (or its isomorph, $D3$) we have for any element $a$ of order $3$, and any element $b$ of order $2$, that:

$bab^{-1} = bab = a^{-1} = a^2$.

Let's see if this is true in $A_4$:

$(1\ 2)(3\ 4)(1\ 2\ 3)(1\ 2)(3\ 4)$ is this:

$1 \to 2 \to 3 \to 4$
$2 \to 1 \to 2 \to 1$
$3 \to 4 \to 4 \to 3$
$4 \to 3 \to 1 \to 2$, that is:

$(1\ 2)(3\ 4)(1\ 2\ 3)(1\ 2)(3\ 4) = (1\ 4\ 2) \neq (1\ 2\ 3)^2$

We can also argue this way:

If $A_4$ has a subgroup $H$ isomorphic to $S_3$, it contains three elements of order $2$, because $S_3$ has three elements of order $2$.

But the only elements of order $2$ in $A_4$ are the 2-transpositions. Since these form a subgroup (along with the identity) $V$ of order 4, we have:

$V \leq H$, but 4 does not divide 6, contradiction.

 

[mathmari]

In both $S_3$ (or its isomorph, $D3$) we have for any element $a$ of order $3$, and any element $b$ of order $2$, that:

$bab^{-1} = bab = a^{-1} = a^2$.

Let's see if this is true in $A_4$:

$(1\ 2)(3\ 4)(1\ 2\ 3)(1\ 2)(3\ 4)$ is this:

$1 \to 2 \to 3 \to 4$
$2 \to 1 \to 2 \to 1$
$3 \to 4 \to 4 \to 3$
$4 \to 3 \to 1 \to 2$, that is:

$(1\ 2)(3\ 4)(1\ 2\ 3)(1\ 2)(3\ 4) = (1\ 4\ 2) \neq (1\ 2\ 3)^2$

So do we have to find one element of $A_4$ so that this equation doesn't hold? (Wondering)

Or do we have to show that this equation is not true for all elements of $A_4$ ? (Wondering)

We can also argue this way:

If $A_4$ has a subgroup $H$ isomorphic to $S_3$, it contains three elements of order $2$, because $S_3$ has three elements of order $2$.

But the only elements of order $2$ in $A_4$ are the 2-transpositions. Since these form a subgroup (along with the identity) $V$ of order 4, we have:

$V \leq H$, but 4 does not divide 6, contradiction.

Why do the 2-transpositions form a subgroup? (Wondering)

 

[Deveno]

So do we have to find one element of $A_4$ so that this equation doesn't hold? (Wondering)

Or do we have to show that this equation is not true for all elements of $A_4$ ? (Wondering)

In $S_3$ and $D_3$ did I not say this holds for ANY element $a$ of order 3, and ANY element $b$ of order 2? To show $A_4$ has a counter-example, we may use any element of order 3, and any element of order 2 we choose.

Why do the 2-transpositions form a subgroup? (Wondering)

That is a very good question...