MHB Verifying Sentences about Groups: Answers and Hints

[mathmari]

Hey!

I have to determine whether the following sentences are correct or not.

1. Any two groups with three elements are isomorphic.
   
2. At any cyclic group, each element is a generator.
   
3. Each cyclic group has at least one non trivial proper subgroup.
   
4. The group $G=\{ 1, i, -1, -i \}$ with respect to the multiplication is isomorphic to $\mathbb{Z}_{4}$.
   
5. Each cyclic group is abelian.
   
6. An abelian group can have a non-abelian subgroup.
   
7. For each $n$ there is an abelian group with order $n$.
   
8. For each $n$ there is a non-abelian group with order $n$.

I have done the following:

1. True, since the multiplicative table is the same with the only difference the symbols.
   
2. False, for example the group $G=\{ 1, -1, i, -i \}$ is cyclic with generator $i$. $<-1>=\{ 1, -1 \} \neq G$, so $-1$ is not a generator of $G$.
   
3. ?
   
4. The multiplication table of the group $G=\{ 1, i, -1, -i \}$ with respect to the multiplication is : $$\begin{bmatrix}
   & | & 1 & i & -1 & -i\\
   - &| &- & - & - & -\\
   1 & |& 1 &i & -1 & -i\\
   i & |& i &-1 &-i &1 \\
   -1 & |&-1 &-i &1 &i \\
   -i & |& -i &1 &i &-1
   \end{bmatrix}$$ and the multiplication table of the group $\mathbb{Z}_4$ is $$\begin{bmatrix}
   & | & 0 & 1 & 2 & 3\\
   - &| &- & - & - & -\\
   0 & |&0 &1 & 2 & 3\\
   1 & |& 1 &2 &3 &0 \\
   2 & |&2 &3 &0 &1 \\
   3 & |& 3 &0 &1 &2
   \end{bmatrix}$$
   
   The tables are the same, the only difference is the symbols ($1 \rightarrow 0, i \rightarrow 1, -1 \rightarrow 2, -i \rightarrow 3$)
   
5. True!
   
6. ?
   
7. ?
   
8. ?

Are my answers correct?? (Wondering)
Could you give me some hints for the ones with the question mark?? (Blush)

 

[Nono713]

I have to determine whether the following sentences are correct or not.

1. Any two groups with three elements are isomorphic.
   
2. At any cyclic group, each element is a generator.
   
3. Each cyclic group has at least one non trivial proper subgroup.
   
4. The group $G=\{ 1, i, -1, -i \}$ with respect to the multiplication is isomorphic to $\mathbb{Z}_{4}$.
   
5. Each cyclic group is abelian.
   
6. An abelian group can have a non-abelian subgroup.
   
7. For each $n$ there is an abelian group with order $n$.
   
8. For each $n$ there is a non-abelian group with order $n$.

1. You are correct. An alternative and more general proof is that any two groups with $p$ elements where $p$ is prime are isomorphic, since a group of prime order is necessarily cyclic.

2. You are correct. A group element is a generator if and only if its order is the order of the group, and in the group you show $-1$ has order $2$, not $4$. Another easy counterexample is the identity, which always has order $1$ by definition (but you might add "non-identity" to the question).

3. Consider a cyclic group of prime order. If it has a subgroup, what could the order of the subgroup be (hint: Lagrange's theorem)? Conclude.

4. You are correct, but you could have also directly invoked the simple isomorphism $\varphi: n \mapsto i^n$ without having to write down multiplication tables.

5. You are correct, this follows trivially from the definition of a cyclic group and the fact that $x^a x^b = x^{a + b} = x^{b + a} = x^b x^a$.

6. Hint: an abelian group is a group where every element commutes with every other. If you have a subgroup of an abelian group, doesn't that still mean every element in the subgroup commutes with every other?

7. Hint: $\mathbb{Z}_2$ is abelian. So is $\mathbb{Z}_3$... and $\mathbb{Z}_4$...

8. What if $n$ is prime? What can you say about groups of prime order? What does that imply?

 

[mathmari]

1. You are correct. An alternative and more general proof is that any two groups with $p$ elements where $p$ is prime are isomorphic, since a group of prime order is necessarily cyclic.

2. You are correct. A group element is a generator if and only if its order is the order of the group, and in the group you show $-1$ has order $2$, not $4$. Another easy counterexample is the identity, which always has order $1$ by definition (but you might add "non-identity" to the question).

3. Consider a cyclic group of prime order. If it has a subgroup, what could the order of the subgroup be (hint: Lagrange's theorem)? Conclude.

4. You are correct, but you could have also directly invoked the simple isomorphism $\varphi: n \mapsto i^n$ without having to write down multiplication tables.

5. You are correct, this follows trivially from the definition of a cyclic group and the fact that $x^a x^b = x^{a + b} = x^{b + a} = x^b x^a$.

6. Hint: an abelian group is a group where every element commutes with every other. If you have a subgroup of an abelian group, doesn't that still mean every element in the subgroup commutes with every other?

7. Hint: $\mathbb{Z}_2$ is abelian. So is $\mathbb{Z}_3$... and $\mathbb{Z}_4$...

8. What if $n$ is prime? What can you say about groups of prime order? What does that imply?

1. So are all cyclic groups with the same order isomorphic??

2. At an additive group does the identity element, $0$, have order $1$, because $0 \cdot 1=0$ or could we also multiply by $0$, $0 \cdot 0=0$??

3. From Lagrange's Theorem we have that the order of the subgroup is either $1$ or $p$ ( where $p$ is the order of the cyclic group). Therefore, the group has no non-trivial proper subgroup.

4. To show that these two groups are isomorphic, do I have to show that $\varphi$ is $1-1$, onto and homomorphism??

5. I understand!

6. I got stuck right now... What does it mean that an element commutes with an other element??

7. So, for every $n$, $\mathbb{Z}_n$ is an abelian group with order $n$??

8. A group $G$ of prime order is isomorphic to $\mathbb{Z}_p$ which is an abelian group. Since isomorphic groups have the same properties, $G$ is also abelian. Therefore, it is not true that there is a non-abelian group with order $n$ for each $n$. Is this correct??

 

[mathbalarka]

1. Yes. Prove it.

2. Multiplying by zero does NOT make sense. $n \cdot a $ is defined to be $a + a + \cdots + a$ for integers $n \in \Bbb N$. What is $0 \cdot e$ supposed to mean? Identity elt $e$ has order $1$.

3. Correct.

4. Yes. Two groups have to be set-isomorphic to be group-isomorphic, but the converse is not necessarily true (take $\Bbb Z_4$ and $V_4$).

6. If $a, b \in G$ satisfies $ab = ba$, $a$ is said to commute with $b$.

7. Yes. Can you prove that?

8. Correct.

 

[mathmari]

1. Yes. Prove it.

2. Multiplying by zero does NOT make sense. $n \cdot a $ is defined to be $a + a + \cdots + a$ for integers $n \in \Bbb N$. What is $0 \cdot e$ supposed to mean? Identity elt $e$ has order $1$.

3. Correct.

4. Yes. Two groups have to be set-isomorphic to be group-isomorphic, but the converse is not necessarily true (take $\Bbb Z_4$ and $V_4$).

6. If $a, b \in G$ satisfies $ab = ba$, $a$ is said to commute with $b$.

7. Yes. Can you prove that?

8. Correct.

1. I tried to prove that all cyclic groups with the same order are isomorphic, as followed:

All cyclic groups of order $n$ are isomorphic to $\mathbb{Z}_n$.
So, let $G_1$ and $G_2$ two cyclic groups of order $n$.
So, we have that $G_1 \cong \mathbb{Z}_n$ and $G_2 \cong \mathbb{Z}_n$.
Therefore, $G_1 \cong G_2$.

Is it correct?? (Wondering)2. I see... If we have a multiplicative cyclic group with generator $a$, is the group then $<a>=\{a^0,a^1,a^2, \dots \}$ or $<a>=\{a^1,a^2, \dots \}$ ??4. $\varphi:n \rightarrow i^n$

1-1 : $x,y \in \mathbb{Z}_4$
We suppose that $x=y$.
$\varphi(x)=i^x=i^y=\varphi(y)$

So, $\varphi$ is $1-1$.

onto : From the definition of $\varphi$, we have that $\varphi$ is onto.

homomorphism : $\varphi(x+y)=i^{x+y}=i^xi^y=\varphi(x)\varphi(y)$

So, $\varphi$ is homomorphism.

Is this right??6. $G$ is abelian $ \Rightarrow ab=ba, \forall a,b \in G$
If $H$ is a subgroup of $G$, then for each $g,h \in H$, we have that $gh=hg$, because $g,h \in G$.

Is is correct??7. Since $\mathbb{Z}_n$ is cyclic, we use the fact that every cyclic group is abelian, right??

 

[Nono713]

1. I tried to prove that all cyclic groups with the same order are isomorphic, as followed:

All cyclic groups of order $n$ are isomorphic to $\mathbb{Z}_n$.
So, let $G_1$ and $G_2$ two cyclic groups of order $n$.
So, we have that $G_1 \cong \mathbb{Z}_n$ and $G_2 \cong \mathbb{Z}_n$.
Therefore, $G_1 \cong G_2$.

Is it correct?? (Wondering)2. I see... If we have a multiplicative cyclic group with generator $a$, is the group then $<a>=\{a^0,a^1,a^2, \dots \}$ or $<a>=\{a^1,a^2, \dots \}$ ??4. $\varphi:n \rightarrow i^n$

1-1 : $x,y \in \mathbb{Z}_4$
We suppose that $x=y$.
$\varphi(x)=i^x=i^y=\varphi(y)$

So, $\varphi$ is $1-1$.

onto : From the definition of $\varphi$, we have that $\varphi$ is onto.

homomorphism : $\varphi(x+y)=i^{x+y}=i^xi^y=\varphi(x)\varphi(y)$

So, $\varphi$ is homomorphism.

Is this right??6. $G$ is abelian $ \Rightarrow ab=ba, \forall a,b \in G$
If $H$ is a subgroup of $G$, then for each $g,h \in H$, we have that $gh=hg$, because $g,h \in G$.

Is is correct??7. Since $\mathbb{Z}_n$ is cyclic, we use the fact that every cyclic group is abelian, right??

1. Yes, that's right. Basically, there is only one kind of cyclic group for every order $n$, and it has a specific structure. And isomorphism is of course an equivalence relation (can you prove that?) so it is transitive (among others).

2. For a finite cyclic group it wouldn't really matter, since eventually you would have $a^k = a^0 = e$ for some positive $k$ (the order of the group) but for infinite cyclic groups there is no such $k$, so you do want to start at $a^0$ because the subgroup must contain the identity element. In fact you must also include negative powers for infinite groups, so the most general expression must be:

$$\langle a \rangle = \left \{ a^k \mid k \in \mathbb{Z} \right \}$$

For instance, consider the additive group of integers with generator $1$ (it doesn't matter that it's not a multiplicative group, it is a group nonetheless). By your definition $1$ would only generate all integers greater than $1$. But it also generates zero and all negative numbers.

4. No, for $\varphi$ to be one-to-one you need to show that $\varphi(x) = \varphi(y) ~ \implies ~ x = y$. The converse is trivially true by definition of a function (if it doesn't hold, something has gone seriously wrong). Here you can show that in complex numbers under multiplication, which is the parent group of $G$, $\varphi(x) = \varphi(y)$ implies $i^x = i^y$, that is, $i^{x - y} = 1$. This happens only when $x - y \equiv 0 \pmod{4}$ since $i \ne 1$, $i^2 \ne 1$, $i^3 \ne 1$, $i^4 = 1$, which precisely means that $x = y$ (in $\mathbb{Z}_4$). So $x = y$ and the function is one-to-one.

To show the function is onto, you need to show that for every $u \in G$ there exists an $x \in \mathbb{Z}_4$ such that $\varphi(x) = u$. You already know this from the fact that the powers of $i$ from $0$ to $3$ are all distinct. So $\varphi$ is onto.

And, yes, your proof that it is an homomorphism is correct, so you are done and it is an isomorphism. This might seem a lot of work for a simple problem like this, but for larger (possibly infinite) groups the direct approach of comparing multiplication tables rarely works.

6. Yes, that is right. The abelian property carries over to subgroups. Note that the converse is not true, a subgroup can be abelian without its parent group being, e.g. many large non-abelian groups have small cyclic subgroups. Can you think of an example?

7. That is one way of going about it. Can you do it without invoking this theorem though? What algebraic structure does $\mathbb{Z}_n$ represent, after all? What can you say about that structure?

 

[Deveno]

Hey!

I have to determine whether the following sentences are correct or not.

1. Any two groups with three elements are isomorphic.
   
2. At any cyclic group, each element is a generator.
   
3. Each cyclic group has at least one non trivial proper subgroup.
   
4. The group $G=\{ 1, i, -1, -i \}$ with respect to the multiplication is isomorphic to $\mathbb{Z}_{4}$.
   
5. Each cyclic group is abelian.
   
6. An abelian group can have a non-abelian subgroup.
   
7. For each $n$ there is an abelian group with order $n$.
   
8. For each $n$ there is a non-abelian group with order $n$.

I have done the following:

1. True, since the multiplicative table is the same with the only difference the symbols.
   
2. False, for example the group $G=\{ 1, -1, i, -i \}$ is cyclic with generator $i$. $<-1>=\{ 1, -1 \} \neq G$, so $-1$ is not a generator of $G$.
   
3. ?
   
4. The multiplication table of the group $G=\{ 1, i, -1, -i \}$ with respect to the multiplication is : $$\begin{bmatrix}
   & | & 1 & i & -1 & -i\\
   - &| &- & - & - & -\\
   1 & |& 1 &i & -1 & -i\\
   i & |& i &-1 &-i &1 \\
   -1 & |&-1 &-i &1 &i \\
   -i & |& -i &1 &i &-1
   \end{bmatrix}$$ and the multiplication table of the group $\mathbb{Z}_4$ is $$\begin{bmatrix}
   & | & 0 & 1 & 2 & 3\\
   - &| &- & - & - & -\\
   0 & |&0 &1 & 2 & 3\\
   1 & |& 1 &2 &3 &0 \\
   2 & |&2 &3 &0 &1 \\
   3 & |& 3 &0 &1 &2
   \end{bmatrix}$$
   
   The tables are the same, the only difference is the symbols ($1 \rightarrow 0, i \rightarrow 1, -1 \rightarrow 2, -i \rightarrow 3$)
   
5. True!
   
6. ?
   
7. ?
   
8. ?

Are my answers correct?? (Wondering)
Could you give me some hints for the ones with the question mark?? (Blush)

1. 3 is prime. So what must be the order of any non-identity element (of which we have 2)?

2. Is the identity EVER a generator for any non-trivial group?

3. Does $\Bbb Z_p$ (the integers mod $p$, for a prime $p$) contain any such groups?

4. Yes, $\phi: \Bbb Z_4 \to \langle i\rangle$ given by $\phi(k) = i^k$ is one such isomorphism (can you find another?).

5. Yes, because in a cyclic group, we ADD the (integer) exponents, and the integers are abelian under addition.

6. Suppose $G$ is abelian, and $H$ is a non-abelian subgroup. We must have $x,y \in H$ with $xy \neq yx$. But $x,y$ are also in $G$. Why is this a contradiction?

7. $\Bbb Z_n$ is such a subgroup.

8. All groups of prime order are not only abelian, but cyclic, as well. Remember this. Thus there is (for example) no non-abelian group of order 3. In fact, the smallest non-abelian group, has order 6 (this is the smallest positive integer that is composite, but not a prime square). The smallest non-abelian group that has prime power order is of order 8.