How do we get a contradiction?

In summary, the author is asking if $D_3$ is also generated by any element of order $2$, together with any element of order $3$. There is no typo in the post.
  • #1
mathmari
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Hey! :eek:

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)
 
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  • #2
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?
 
  • #3
Deveno said:
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.
Deveno said:
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)
 
  • #4
mathmari said:
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).
 
  • #5
Deveno said:
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)
 
  • #6
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 $3$.
 
  • #7
Deveno said:
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)
Deveno said:
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)
 
  • #8
mathmari said:
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).
 
  • #9
Deveno said:
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)
 
  • #10
mathmari said:
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.
 
  • #11
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)
 
  • #12
mathmari said:
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).
 
  • #13
Deveno said:
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)

Deveno said:
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)
 
  • #14
mathmari said:
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.
 
  • #15
Deveno said:
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)
Deveno said:
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)
 
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  • #16
mathmari said:
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...
 

Related to How do we get a contradiction?

1. What is a contradiction?

A contradiction is a statement or situation that is logically opposite or inconsistent with another statement or situation. It is often seen as a logical error or fallacy.

2. Why do we want to find a contradiction?

Finding a contradiction can help us identify flaws or inconsistencies in our reasoning or arguments. It can also lead to new insights or discoveries in a field of study.

3. How do we look for a contradiction?

One way to look for a contradiction is to carefully examine the premises and conclusions of an argument or statement. We can also test the logic and validity of the argument to see if it holds up under scrutiny.

4. What happens if we find a contradiction?

If we find a contradiction, it means that there is a logical error or inconsistency in the argument or statement. This may require us to revise our reasoning or conclusions and potentially lead to new discoveries or insights.

5. Can a contradiction be resolved?

Yes, a contradiction can be resolved by identifying and addressing the error or inconsistency. This may involve revising our reasoning, gathering more evidence, or seeking alternative explanations.

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