1. Isomorphism

I have five problems from my text that I did exactly the same way. Thus I have reason to believe that I am doing something wrong.

The problem is showing that two groups are not isomorphic. To give you the flavor of what I am doing consider the following:
Prove that the additive groups $\mathbb{Z}$ and $\mathbb{Q}$ are not isomorphic.
It is clear to me that we can construct a function $\phi : \mathbb{Z} \to \mathbb{Q}$ that is surjective. The original reason I claimed that the function can't be injective is because $\mathbb{Q}$ is a two parameter group and $\mathbb{Z}$ is only a one parameter group, so we cannot construct a 1 - 1 function, therefore the groups cannot be isomorphic. I don't see this reasoning to be correct any more as I've been applying this over and over in the exercises. (I don't see any reason the text would hammer the point over and over if the reasoning is this "trivial.")

So how would I show that these groups are not isomorphic?

-Dan

2. Re: Isomorphism

Actually the extension of the Cantor pairing set is an injection between the integers and the rationals. Now I'm in real trouble (pun not intended!) because this would represent a bijection. So more !

-Dan

3. Re: Isomorphism

Use the properties of an isomorphism. Suppose BWOC that $\phi:\mathbb{Z} \to \mathbb{Q}$ is an isomorphism. Then $\phi(0) = \phi(0+0) = \phi(0) + \phi(0)$. Subtract $\phi(0)$ from both sides and you find $\phi(0) = 0$. Since this is an isomorphism, $\phi(1) \neq 0$. But then,

\begin{align*}1 & = \phi^{-1}\left(\phi(1) \right) \\ & = \phi^{-1}\left(\dfrac{1}{2}\phi(1) + \dfrac{1}{2}\phi(1)\right) \\ & = \phi^{-1}\left(\dfrac{1}{2}\phi(1)\right)+\phi^{-1}\left(\dfrac{1}{2}\phi(1)\right) \\ & = 2\phi^{-1}\left(\dfrac{1}{2}\phi(1)\right)\end{align*}

Solving for $\phi^{-1}\left(\dfrac{1}{2}\phi(1)\right)$, we find $\phi^{-1}\left(\dfrac{1}{2}\phi(1)\right) = \dfrac{1}{2} \notin \mathbb{Z}$, contradicting the assumption that $\phi$ is an isomorphism.

4. Re: Isomorphism

Originally Posted by SlipEternal
Use the properties of an isomorphism. Suppose BWOC that $\phi:\mathbb{Z} \to \mathbb{Q}$ is an isomorphism. Then $\phi(0) = \phi(0+0) = \phi(0) + \phi(0)$. Subtract $\phi(0)$ from both sides and you find $\phi(0) = 0$. Since this is an isomorphism, $\phi(1) \neq 0$. But then,

\begin{align*}1 & = \phi^{-1}\left(\phi(1) \right) \\ & = \phi^{-1}\left(\dfrac{1}{2}\phi(1) + \dfrac{1}{2}\phi(1)\right) \\ & = \phi^{-1}\left(\dfrac{1}{2}\phi(1)\right)+\phi^{-1}\left(\dfrac{1}{2}\phi(1)\right) \\ & = 2\phi^{-1}\left(\dfrac{1}{2}\phi(1)\right)\end{align*}

Solving for $\phi^{-1}\left(\dfrac{1}{2}\phi(1)\right)$, we find $\phi^{-1}\left(\dfrac{1}{2}\phi(1)\right) = \dfrac{1}{2} \notin \mathbb{Z}$, contradicting the assumption that $\phi$ is an isomorphism.
I see and understand, thank you! But how would you know to look for something like that? Or is it one of those "you have to see it done first" sort of things?

I'll look at the other problems in the text and see if I can construct some similar kinds of arguments.

-Dan

5. Re: Isomorphism

The idea of an isomorphism is that two groups have the same additive structure. So I immediately thought to start with an assumed isomorphism from $\mathbb{Q}\to \mathbb{Z}$ and check out what happens when I send one half through. If I send it through twice, I should get 1. But, I have no idea where one half is being sent. So, I needed to know which rational number is sent to one. So, it seemed more natural to use a map from $\mathbb{Z}\to \mathbb{Q}$ and just send 1 through. Then I know exactly which rational number is mapped from 1.

6. Re: Isomorphism

Dan,
Here's maybe a more "group theoretic" proof.

Assume the two are isomorphic. Since $\mathbb{Z}$ is cyclic, $\mathbb{Q}$ is cyclic. Let ${a\over b}$ be a generator of $\mathbb{Q}$. Then every element of $\mathbb{Q}$ is of the form ${na\over b}$ for some integer $n$. In particular, ${na\over b}={1\over b}$ for some $n\in\mathbb{Z}$. So $na=1$ and thus $a=\pm1$. In any case then ${1\over b}$ is a generator of $\mathbb{Q}$. (If x is a generator of a cyclic group, so also is x inverse). Pick any prime p which does not divide b. Then for some $m\in\mathbb{Z}$, ${m\over b}={1\over p}$ or mp=b, contradicting the choice of p.

Addendum: Without seeing the other exercises, once you dismiss the possibility that the two groups in question have different cardinalities, I would look for group properties. Examples: cyclic, abelian, numbers of elements of a given order, etc.

Example:
Suppose G is a group of order 12 that has a subgroup H of order 6. Then G is not isomorphic to A4, the alternating group on 4 letters.
Proof. H has a subgroup of order 3, which is then normal in G. However, A4 has no normal subgroup of order 3. QED

7. Re: Isomorphism

Originally Posted by johng
Dan,
Here's maybe a more "group theoretic" proof.

Assume the two are isomorphic. Since $\mathbb{Z}$ is cyclic, $\mathbb{Q}$ is cyclic. Let ${a\over b}$ be a generator of $\mathbb{Q}$. Then every element of $\mathbb{Q}$ is of the form ${na\over b}$ for some integer $n$. In particular, ${na\over b}={1\over b}$ for some $n\in\mathbb{Z}$. So $na=1$ and thus $a=\pm1$. In any case then ${1\over b}$ is a generator of $\mathbb{Q}$. (If x is a generator of a cyclic group, so also is x inverse). Pick any prime p which does not divide b. Then for some $m\in\mathbb{Z}$, ${m\over b}={1\over p}$ or mp=b, contradicting the choice of p.

Addendum: Without seeing the other exercises, once you dismiss the possibility that the two groups in question have different cardinalities, I would look for group properties. Examples: cyclic, abelian, numbers of elements of a given order, etc.

Example:
Suppose G is a group of order 12 that has a subgroup H of order 6. Then G is not isomorphic to A4, the alternating group on 4 letters.
Proof. H has a subgroup of order 3, which is then normal in G. However, A4 has no normal subgroup of order 3. QED
I like the generality of this one, thank you as well! However it is a bit outside what the text might expect at this point.

-Dan

8. Re: Isomorphism

Okay. I'm going go down the list. Please correct me if I get any wrong. I need to show that each associated groups are not isomorphic.

Assume each function $\phi$ exists and is an isomorphism between the mentioned groups. (The site dictionary has "isomorphic" but not "isomorphism"??)

1) $\phi : \mathbb{R} - \{ 0 \} \to \mathbb{C} - \{ 0 \}$ as multiplicative groups.
This is not an isomorphism because the element i in $\mathbb{C} - \{ 0 \}$ has order 4 and no elements of $\mathbb{R} - \{ 0 \}$ have order 4.

2) $\phi: \mathbb{R} \to \mathbb{Q}$ as additive groups.
This is not an isomorphism as the groups have different cardinalities. (What the heck? We really need to upgrade the site dictionary!)

3) $\phi : D_8 \to Q_8$
This is not an isomorphism as $D_8$ has 4 elements of order 4 and $Q_8$ has 6 elements of order 4.

4) $\phi : S_n \to S_m$ where $n \neq m$
This is not an isomorphism because $|S_n| \neq |S_m|$.

5) $\phi : D_{24} \to S_4$
This is not an isomporphism as $D_{24}$ only has three elements of order 2 and $S_4$ has 9 elements of order 2. (Did I count that right?)

Thanks!

-Dan

PS Just for reference, the notation $\mathbb{R} - \{ 0 \}$ is the same as $\mathbb{R} \setminus \{ 0 \}$. They both mean "take 0 from the set," yes?

9. Re: Isomorphism

Dan,
1 ,2 and 4 look good.

3. $D_8=\{1,x,x^2,x^3,y,xy,x^2y,x^3y\}$, here $x^4=y^2=1$ and $yxy=x^{-1}$. So the only elements of order 4 are $x$ and $x^3$; $x^2, y$ have order 2 and $(x^iy)(x^iy)=x^ix^{-i}=1$

$Q_8$ formally has the same elements as $D_8$, but $x^4=1, y^2=x^2$ and $y{-1}xy=x^{-1}$. I leave it to you to compute exactly 4 elements of order 4.

So you have the right idea, but your counts are off.

5. $D_{24}$ has 13 elements of order 2, whereas $S_4$ has 9 elements of order 2. This is a lot of computation, but it does show they are not isomorphic. Maybe easier: convince your self that the order of a product of disjoint cycles in a permutation group has order the least common multiple of the cycle lengths. So the maximum order of an element of $S_4$ is 4 (a 4 cycle), whereas $D_{24}$ has an element of order 12.

By the way, I think the usual symbol for a dihedral group is $D_n$, the dihedral group with 2n elements (at least it used to be). Also the quaternion group (your notation $Q_8$) is usually denoted just Q. The generalized quaternion group has 2n elements; I don't remember any standard notation for this generalized quaternion group.

Generally, a good job.

10. Re: Isomorphism

Originally Posted by johng
Dan,
1 ,2 and 4 look good.

3. $D_8=\{1,x,x^2,x^3,y,xy,x^2y,x^3y\}$, here $x^4=y^2=1$ and $yxy=x^{-1}$. So the only elements of order 4 are $x$ and $x^3$; $x^2, y$ have order 2 and $(x^iy)(x^iy)=x^ix^{-i}=1$

$Q_8$ formally has the same elements as $D_8$, but $x^4=1, y^2=x^2$ and $y{-1}xy=x^{-1}$. I leave it to you to compute exactly 4 elements of order 4.

So you have the right idea, but your counts are off.

5. $D_{24}$ has 13 elements of order 2, whereas $S_4$ has 9 elements of order 2. This is a lot of computation, but it does show they are not isomorphic. Maybe easier: convince your self that the order of a product of disjoint cycles in a permutation group has order the least common multiple of the cycle lengths. So the maximum order of an element of $S_4$ is 4 (a 4 cycle), whereas $D_{24}$ has an element of order 12.

By the way, I think the usual symbol for a dihedral group is $D_n$, the dihedral group with 2n elements (at least it used to be). Also the quaternion group (your notation $Q_8$) is usually denoted just Q. The generalized quaternion group has 2n elements; I don't remember any standard notation for this generalized quaternion group.

Generally, a good job.
Thank you. Yes, Q_8 in my text is the quaternions and the notation for the dihedral groups is D_2n, rather than D_n.

For D_{24}, calling r the rotation element and s the inverse element, how do we get 13 elements of order 2? I'm only finding r^12, s, and sr^12 as order 2. Which other elements are there?

Thanks!

-Dan

11. Re: Isomorphism

Whoops! I meant r^6, s, and sr^6 for order 2 elements in D_24.

-Dan

12. Re: Isomorphism

Dan,
You need to know for any group G with x and y in G, $y^{-1}x^my=(y^{-1}xy)^m$ for any integer m. This is because conjugation is an isomorphism, or you can prove it directly.
So in $D_{24}$, if r is the rotation (order 12) and s is the reflection,

$sr^msr^m=s^{-1}r^msr^m=(s^{-1}rs)^mr^m=(r^{-1})^mr^m=r^{-m}r^m=1$

So you have 12 elements srm, (m ranging from 0 to 11) or order 2. With r6, a total of 13 elements of order 2.

Geometrically, draw the regular 12-gon symmetrically about the y axis; let r be rotation by 30 degrees and s the reflection about the y axis. Think of first reflecting, then rotating by any multiple of 30. (or in the other order). If you do this twice, you're back to where you started.

13. Re: Isomorphism

Originally Posted by johng
Dan,
You need to know for any group G with x and y in G, $y^{-1}x^my=(y^{-1}xy)^m$ for any integer m. This is because conjugation is an isomorphism, or you can prove it directly.
So in $D_{24}$, if r is the rotation (order 12) and s is the reflection,

$sr^msr^m=s^{-1}r^msr^m=(s^{-1}rs)^mr^m=(r^{-1})^mr^m=r^{-m}r^m=1$

So you have 12 elements srm, (m ranging from 0 to 11) or order 2. With r6, a total of 13 elements of order 2.

Geometrically, draw the regular 12-gon symmetrically about the y axis; let r be rotation by 30 degrees and s the reflection about the y axis. Think of first reflecting, then rotating by any multiple of 30. (or in the other order). If you do this twice, you're back to where you started.
Ah well. Oops! Thanks for the catch!

-Dan