0
$\begingroup$

I want to prove that if $G = \{g_1, g_2, ..., g_n\}$ is a finite group and $h \in G$, then $G = \{hg_1,hg_2, ..., hg_n\} $.

My attempt:

Let $hg_i = hg_j = g_k \quad\quad 1 \leq i,j,k \leq n$.

Then:

$g_i = g_j$

By contraposition, we find that $gi \neq g_j \Rightarrow hg_i \neq hg_j$. Thus, all elements in $\{hg_1,hg_2, ..., hg_n\}$ are different. As this is a subset of $G$, with as many elements as in $G$, we deduce that $G = \{hg_1,hg_2, ..., hg_n\}$

Can someone verify whether this is correct?

  • 3
    You might add half a line as to *why* $hg_i=hg_j=g_k$ implies $g_i=g_j$. -- For an alternative proof, note that the map $G\to G$, $x\mapsto h^{-1}x$ is a right and left inverse inverse map to $x\mapsto hx$, hence the latter is injective and surjective. It may be interesting to note that finiteness of $G$ is not needed for the proof. – 2017-02-24
  • 0
    Although I think your proof looks technically correct it's kind of weird. Notice that your second set is the image of $G$ under $g\mapsto hg$ for $g\in G$. Can you think of an inverse map? – 2017-02-24
  • 0
    @Hagen von Eitzen We can multiply both sides with the inverse of h (which exists by definition of group) to the left, then use associativity and definitions of neutral element, inverse element to get what we want. – 2017-02-24
  • 0
    @RobertChamberlain These are just the basics for me. We did not study isomorphisms and other functions yet, but I think the inverse function would be $g \mapsto h^{-1}g$ – 2017-02-24
  • 1
    @Math_QED you're exactly right (though I realise Hagen did get his comment down before me). As he explained, the existence of a left and right inverse of $g\mapsto hg$ implies that it is bijective – 2017-02-24

1 Answers 1

0

The approach in my original post is correct, but this is a more intuitive way.

Call $H = \{hg_1, \dots,hg_n\}$

Define a mapping $\phi: G\rightarrow H: g \mapsto hg.$ We prove that this mapping is bijective:

Surjective:

Let $y \in H$. Then there is a $g_i \in G$ such that $y = hg_i$. Because $\phi(g_i) = y$, the mapping is surjective.

Injective:

Let $g_i, g_j \in G$ with $\phi(g_i) = \phi(g_j)$. Then $hg_i = hg_j$ and because $h,g_i,g_j$ are elements of the group G, we can cancel the $h$ on both sides, obtaining that $g_i = g_j$. Hence, $\phi$ is injective.

Now, we have shown that this mapping is bijective. Because any bijective mapping between $2$ finite sets implies that both sets contain the same amount of elements, we are done.

Notice that we also could construct the inverse function (as mentioned in the comments) to show bijectivity.