Please review my proof

[Boi]

Task says the following: If $G$ is a group in which $(a\cdot b)^i=a^i \cdot b^i$ for three consecutive integers $i$ for all $a, b \in G$ show that $G$ is abelian.
This is my proof: first I want to show that there is no need to consider cases where one of $i$ is negative. We have $$(a\cdot b)^{-k}=a^{-k} \cdot b^{-k}$$ $$(a \cdot b)^k=(a^{-k} \cdot b^{-k})^{-1}=(b^{-k})^{-1} \cdot (a^{-k})^{-1}=b^k \cdot a^k.$$ Then, by multiplying by $b$ on the left and by $a$ on the right $$b \cdot (a \cdot b)^k \cdot a=b^{k+1} \cdot a^{k+1}$$ and regrouping the terms $$b \cdot (a \cdot b)^k \cdot a=b \cdot(a\cdot b) \cdot ... \cdot(a \cdot b) \cdot a=(b\cdot a) \cdot (b \cdot a) \cdot ... \cdot (b \cdot a)=(b \cdot a)^{k+1}$$ we have $$(b \cdot a)^{k+1}=b^{k+1} \cdot a^{k+1}.$$ So, we now have exactly the same (just with changed labels) relation but with a positive exponent $k+1$. Before moving on, I only need to mention that relation $(a \cdot b)^m=a^m \cdot b^m$ implies $(b \cdot a)^{m-1}=a^{m-1} \cdot b^{m-1}$, since $$(a \cdot b)^m=a^m \cdot b^m$$ $$(a \cdot b) \cdot ... \cdot (a \cdot b)=a^m \cdot b^m$$ canceling $a$ from the left and $b$ from the right and regrouping the terms we have $$(b \cdot a)^{m-1} =a^{m-1} \cdot b^{m-1}.$$ Now, let n denote the smallest of the successive integers. Relations $(a \cdot b)^{n+1}= a^{n+1} \cdot b^{n+1}$ and $(a \cdot b)^{n+2}= a^{n+2} \cdot b^{n+2}$ imply $(b \cdot a)^{n} =a^{n} \cdot b^{n}$ and $(b \cdot a)^{n+1} =a^{n+1} \cdot b^{n+1}$ respectively. With that, I conclude $$(a \cdot b)^{n+1}=a^{n+1} \cdot b^{n+1}=(b \cdot a)^{n+1}$$ $$(a \cdot b) \cdot (a \cdot b)^n=(b \cdot a) \cdot (b \cdot a)^n.$$ Because $(b \cdot a)^{n} =a^{n} \cdot b^{n}= (a \cdot b)^n$ I can cancel it from the right to get $a \cdot b = b \cdot a$ which means that $G$ is abelian. Q.E.D.
I kindly ask if somebody could tell me if this proof is correct. I also would like to know how you check your own prooves when you need to since I don't want to flood this forum with my posts. Thank you for reading.

 

[cbarker12]

first I want to show that there is no need to consider cases where one of iii is negative. We have (a⋅b)−k=a−k⋅b−k(a\cdot b)^{-k}=a^{-k} \cdot b^{-k}(a⋅b)−k=a−k⋅b−k (a⋅b)k=(a−k⋅b−k)−1=(b−k)−1⋅(a−k)−1=bk⋅ak.(a \cdot b)^k=(a^{-k} \cdot b^{-k})^{-1}=(b^{-k})^{-1} \cdot (a^{-k})^{-1}=b^k \cdot a^k.(a⋅b)k=(a−k⋅b−k)−1=(b−k)−1⋅(a−k)−1=bk⋅ak.

What does $k$ stands for and is $k$ arbitrary or not?

Then, by multiplying by bbb on the left and by aaa on the right b⋅(a⋅b)k⋅a=bk+1⋅ak+1b \cdot (a \cdot b)^k \cdot a=b^{k+1} \cdot a^{k+1}b⋅(a⋅b)k⋅a=bk+1⋅ak+1 and regrouping the terms b⋅(a⋅b)k⋅a=b⋅(a⋅b)⋅...⋅(a⋅b)⋅a=(b⋅a)⋅(b⋅a)⋅...⋅(b⋅a)=(b⋅a)k+1b \cdot (a \cdot b)^k \cdot a=b \cdot(a\cdot b) \cdot ... \cdot(a \cdot b) \cdot a=(b\cdot a) \cdot (b \cdot a) \cdot ... \cdot (b \cdot a)=(b \cdot a)^{k+1}b⋅(a⋅b)k⋅a=b⋅(a⋅b)⋅...⋅(a⋅b)⋅a=(b⋅a)⋅(b⋅a)⋅...⋅(b⋅a)=(b⋅a)k+1

I recommended that you use associative property from the group $G$.

, I only need to mention that relation (a⋅b)m=am⋅bm(a \cdot b)^m=a^m \cdot b^m(a⋅b)m=am⋅bm implies (b⋅a)m−1=am−1⋅bm−1(b \cdot a)^{m-1}=a^{m-1} \cdot b^{m-1}(b⋅a)m−1=am−1⋅bm−1, since (a⋅b)m=am⋅bm(a \cdot b)^m=a^m \cdot b^m(a⋅b)m=am⋅bm (a⋅b)⋅...⋅(a⋅b)=am⋅bm(a \cdot b) \cdot ... \cdot (a \cdot b)=a^m \cdot b^m(a⋅b)⋅...⋅(a⋅b)=am⋅bm canceling aaa from the left and bbb from the right and regrouping the terms we have (b⋅a)m−1=am−1⋅bm−1.(b \cdot a)^{m-1} =a^{m-1} \cdot b^{m-1}.(b⋅a)m−1=am−1⋅bm−1.

Why does $a^m b^m=a^{m-1}b^{m-1}$ and what does $m$ stands for?

Now, let n denote the smallest of the successive integers. Relations (a⋅b)n+1=an+1⋅bn+1(a \cdot b)^{n+1}= a^{n+1} \cdot b^{n+1}(a⋅b)n+1=an+1⋅bn+1 and (a⋅b)n+2=an+2⋅bn+2(a \cdot b)^{n+2}= a^{n+2} \cdot b^{n+2}(a⋅b)n+2=an+2⋅bn+2 imply (b⋅a)n=an⋅bn(b \cdot a)^{n} =a^{n} \cdot b^{n}(b⋅a)n=an⋅bn and (b⋅a)n+1=an+1⋅bn+1(b \cdot a)^{n+1} =a^{n+1} \cdot b^{n+1}(b⋅a)n+1=an+1⋅bn+1 respectively. With that, I conclude (a⋅b)n+1=an+1⋅bn+1=(b⋅a)n+1(a \cdot b)^{n+1}=a^{n+1} \cdot b^{n+1}=(b \cdot a)^{n+1}(a⋅b)n+1=an+1⋅bn+1=(b⋅a)n+1 (a⋅b)⋅(a⋅b)n=(b⋅a)⋅(b⋅a)n.(a \cdot b) \cdot (a \cdot b)^n=(b \cdot a) \cdot (b \cdot a)^n. (a⋅b)⋅(a⋅b)n=(b⋅a)⋅(b⋅a)n. Because (b⋅a)n=an⋅bn=(a⋅b)n(b \cdot a)^{n} =a^{n} \cdot b^{n}= (a \cdot b)^n(b⋅a)n=an⋅bn=(a⋅b)n I can cancel it from the right to get a⋅b=b⋅aa \cdot b = b \cdot aa⋅b=b⋅a which means that G is abelian. Q.E.D.

Why does that lead to $G$ being an abelian group?

In General, I think your proof is correct. I recommend you to rewrite the proof in complete sentences, which means to add periods and commas as well "Relations" means what. I recommend that you also avoid using "let" unless you have unambiguous assumption. Define the variables determine whether it is existential or universial values.

cbarker12

 

[Boi]

What does $k$ stands for and is $k$ arbitrary or not?

I recommended that you use associative property from the group $G$.

Why does $a^m b^m=a^{m-1}b^{m-1}$ and what does $m$ stands for?



Why does that lead to $G$ being an abelian group?

In General, I think your proof is correct. I recommend you to rewrite the proof in complete sentences, which means to add periods and commas as well "Relations" means what. I recommend that you also avoid using "let" unless you have unambiguous assumption. Define the variables determine whether it is existential or universial values.

cbarker12

Thank you for your reply!
$k$ is just an arbitrary positive integer.
Isn't that what I used there? Without associativity, I wouldn't be able to just open up brackets like that. Or are you saying that I should've stated that I'm using it?
$m$ is some positive integer.
Because, as stated in task, $(a \cdot b)^n=a^n \cdot b^n$, $(a \cdot b)^{n+1}=a^{n+1} \cdot b^{n+1}$, $(a \cdot b)^{n+2}=a^{n+2} \cdot b^{n+2}$ is true for any two elements $a, b \in G$. Hence, everything concluded from these relations should also apply for any two elements of the group. Oh, and by "relations" I meant thngs like $a \cdot b =b^{-1} \cdot a$, or $a \cdot b \cdot a^{-1}=b^4$. Like, an identity which holds for any two elements of the group. I'm sorry if this is not the right term to use.

 

[cbarker12]

Thank you for your reply!
$k$ is just an arbitrary positive integer.
Isn't that what I used there? Without associativity, I wouldn't be able to just open up brackets like that. Or are you saying that I should've stated that I'm using it?
$m$ is some positive integer.
Because, as stated in task, $(a \cdot b)^n=a^n \cdot b^n$, $(a \cdot b)^{n+1}=a^{n+1} \cdot b^{n+1}$, $(a \cdot b)^{n+2}=a^{n+2} \cdot b^{n+2}$ is true for any two elements $a, b \in G$. Hence, everything concluded from these relations should also apply for any two elements of the group. Oh, and by "relations" I meant thngs like $a \cdot b =b^{-1} \cdot a$, or $a \cdot b \cdot a^{-1}=b^4$. Like, an identity which holds for any two elements of the group. I'm sorry if this is not the right term to use.

Explicitly define the symbols for each variable in your proof. Assert the utilization of the associative property from the group. Instead of using a relation, we will employ the following identity. Initially, it's okay to use incorrect terminology, but as the draft progresses, it's important to use more precise terms.

 

[Boi]

I see, Let me try again.
Task: If $G$ is a group in which $( a \cdot b)^i = a^i \cdot b^i$ for three consecutive integers $i$ for all $a, b \in G$ show that $G$ is abelian.
Proof: before prooving what's needed, I want to proove 2 statements. Statement 1: if $G$ is a group, $k$ is a positive integer and for all $a, b \in G$ the following identity holds $(a \cdot b)^{-k}=a^{-k} \cdot b^{-k}$, then identity $(b \cdot a)^{k+1}=b^{k+1} \cdot a^{k+1}$ also holds for any $a, b \in G$. Proof of statement 1: $$(a \cdot b)^{-k}=a^{-k} \cdot b^{-k}$$$$(a \cdot b )^k =(a^{-k} \cdot b^{-k})^{-1}=(b^{-k})^{-1} \cdot(a^{-k})^{-1}=b^k \cdot a^k$$ then, by multiplying by $b$ on the left and by $a$ on the right, we obtain $$b \cdot (a \cdot b)^k \cdot a= b^{k+1} \cdot a^{k+1}.$$ Now, because $a, b \in G$ and $G$ is a group we can exploit associativity by freely moving the brackets around $$b \cdot (a \cdot b)^k \cdot a= b \cdot (a \cdot b)\cdot... \cdot(a \cdot b) \cdot a= (b\cdot a) \cdot ...\cdot (b \cdot a)=(b \cdot a)^{k+1}=b^{k+1} \cdot a^{k+1}.$$Statement 2: if $G$ is a group, $k$ is a positive integer and for all $a, b \in G$ identity $(a \cdot b)^k =a^k \cdot b^k$ holds, then identity $(b \cdot a)^{k-1}=a^{k-1} \cdot b^{k-1}$ also holds for any two elements in $G$. Proof of statement 2: I start by using associative property of multiplication in $G$ $$(a\cdot b)^k=(a\cdot b)\cdot ... \cdot(a\cdot b)= a \cdot (b\cdot a)\cdot ... \cdot (b \cdot a) \cdot b = a \cdot (b \cdot a)^{k-1} \cdot b$$ and then canceling $a$ and $b$ $$a \cdot (b \cdot a)^{k-1} \cdot b=a^k \cdot b^k$$ $$(b \cdot a)^{k-1}=a^{k-1} \cdot b^{k-1}.$$So, onto the task. I will denote the smallest of $i$'s with $n$. Because of statement 1 I can freely assume that $n \ge 0$ ( if $n$ was negative we would translate all of the identities involving negative exponents into identities involving only positive exponents and then work with them). Identities $(a \cdot b)^{n+1}=a^{n+1} \cdot b^{n+1}$ and $(a \cdot b)^{n+2}= a^{n+2} \cdot b^{n+2}$ imply identities $(b \cdot a)^n =a^n \cdot b^n$ and $(b \cdot a)^{n+1}= a^{n+1} \cdot b^{n+1}$ (because of statement 2). So $$(a \cdot b)^{n+1}=a^{n+1} \cdot b^{n+1}=(b \cdot a)^{n+1}$$$$(a \cdot b) \cdot (a\cdot b)^n=(b \cdot a) \cdot (b \cdot a)^n$$ but $(a \cdot b)^n =a^n \cdot b^n=(b \cdot a)^n$. This means we can cancel $(a \cdot b)^n$ and $(b \cdot a)^n$ from both sides to get $$a \cdot b= b \cdot a.$$ Q.E.D.
I hope this one is better

 

[Steven G]

$$(a \cdot b )^k =(a^{-k} \cdot b^{-k})^{-1}=(b^{-k})^{-1} \cdot(a^{-k})^{-1}=b^k \cdot a^k$$

WHY is the first equality sign valid??

 

[Boi]

WHY is the first equality sign valid??

Oh, well that's because I assumed that $(a \cdot b)^{-k}$ equals $a^{-k} \cdot b^{-k}$ which means that their inverses must also be equal, hence the first equality sign.

 

[Steven G]

Sorry for the late reply but I was out of my country.
I am not sure what you are really saying.
Your 1st equality states that (ab)^k= (a^-kb^-k)^-1. Why are you assuming this to be true??

 

[Boi]

Sorry for the late reply but I was out of my country.
I am not sure what you are really saying.
Your 1st equality states that (ab)^k= (a^-kb^-k)^-1. Why are you assuming this to be true??

Inverse of $(a \cdot b) ^{-k}$ is $(a \cdot b) ^k$ and inverse of $a^{-k} \cdot b^{-k}$ is $(a^{-k} \cdot b^{-k}) ^{-1}$. They are equal because $(a \cdot b) ^{-k}$ and $a^{-k} \cdot b^{-k}$ are equal.