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Let $(G,\cdot)$ be a group. Consider a set $H\neq \emptyset$ such that there exists a bijective function $f:G\to H$ with the property that $f(x\cdot y)=f(x)\cdot f(y)$,$\forall x\in G$. Is it true then that $(H,\cdot)$ is also a group?

I am not able neither to prove this nor disprove it, but I will provide how I came up with this.

I started from the well-known fact that if $(G,\cdot)$ is a group and we have a set $M\neq \emptyset$ such that there exists a bijective function $f:G\to M$, then the following assertions are true:

a) There exists a unique binary operation $*$ such that $f(x\cdot y)=f(x)*f(y)$,$\forall x,y\in G$, and this operation is $\forall a,b \in M, a*b=f(f^{-1}(a)\cdot f^{-1}(b))$.

b)$(M,*)$ is a group isomorphic to $(G,\cdot)$.

After seeing this result, I wanted to find out if there is some way to have the same binary operation on both sets instead of defining a new one. This is how I came up with the result from the beginning of my post. Intuitively, it seems true, because $f$ is something like a group isomorphism and I think that it should "transport" $(G,\cdot)$'s structure to $(H,\cdot)$.

However, I can't prove this.

Should the result I thought up be false, please show me under what conditions we may obtain what I want.

Shaun
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JoMath
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    $f(x) \cdot f(y)$ hasn't been defined yet; you also need to specify a multiplication on $H$ even if you haven't required any axioms of it. – Qiaochu Yuan Sep 16 '19 at 19:22
  • @QiaochuYuan right, I didn't pay attention and I forgot to mention this, even though I think it was pretty obvious that I was defining the same multiplication on $H$ too. Nevertheless, you are right, and I thank you for your spot on observation. – JoMath Sep 16 '19 at 19:41
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    @JoMath Your comment about it being "pretty obvious" is incorrect, in the sense that "the same multiplication on $H$ too" doesn't make sense! I mean, you can attempt to define a multiplication on $H$ which somehow corresponds to the multiplication on $G$, and in fact this is possible. Understand how to do this, and you will be in a much better place to answer this question. – user1729 Sep 16 '19 at 19:54
  • @user1729 Let me try to make clearer what I said : by "the same multiplication on $H$ too" I mean that $x \cdot y$ for $x,y\in H$ has the same formula as $x\cdot y$, for $x,y\in G$. Of course, this implies the assumption that $\cdot$ is a binary operation on $H$ too. I think that this is all right, and this is why I said that it looked "pretty obvious", given that we made the assumption that I mentioned. – JoMath Sep 16 '19 at 20:30
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    @JoMath ...but $x$ and $y$ are not elements of $H$. You need to say something like "define multiplication $$ on $H$ as $h_1h_2:=f(f^{-1}(h_1)\cdot f^{-1}(h_2))$ for $h_1, h_2\in H$". – user1729 Sep 17 '19 at 06:32
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    @user1729 Now I finally understand this. My confusion stemmed from the fact that if for instance we had $(G,\cdot)=(\mathbb{C}\times\mathbb{C},\cdot)$ and $H$ were a group of matrices, I was under the impression that the product of matrices was the same operation as the product of numbers in $\mathbb{C}\times\mathbb{C}$, but this is not true, it is just the same notation for different things. Thank you for your help, I wouldn't have figured this out if you hadn't insisted on this subtle detail ! – JoMath Sep 17 '19 at 15:00

1 Answers1

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If $f$ is bijective, then $\forall y \in H$, exists a unique $x \in G$ such that $f(x)=y$. By property of $f$, we have:

i) $\bar{e}=f(e)$ is a neutral element of $H$ (where $e$ is a neutral element of $G$). In fact, $\forall y \in H$: $$ y \cdot \bar{e} = f(x)\cdot f(e) = f(x \cdot e) = f(e) = \bar{e} $$

ii) If $y_1,y_2,y_3 \in H$, then: $$ y_1 \cdot (y_2 \cdot y_3) = f(x_1) \cdot (f(x_2) \cdot f(x_3)) = f(x_1)\cdot f(x_2\cdot x_3) = f(x_1\cdot(x_2\cdot x_3)) = f((x_1 \cdot x_2)\cdot x_3) = (y_1\cdot y_2)\cdot y_3 $$

iii) For all $y \in H$, exists a unique $x \in G$ such that $f(x)=y$. Then, $y^{-1}=f(x^{-1})$, where $x^{-1}$ is a inverse of $x$. In fact:

$$ y \cdot y^{-1} = f(x)\cdot f(x^{-1}) = f(x\cdot x^{-1}) = f(e) = \bar{e} $$

Therefore, $(H,\cdot)$ is a group.

Rodolfo Ferreira de Oliveira
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