One sentence from Amstrong's Group and Symmetry wrote the following to prove a group of order 6 is isomorphic to $\mathbf{Z}_6$:

The right cosets $\langle x\rangle$, $\langle x\rangle y$ give 6 elements $e$, $x$, $x^2$, $y$, $xy$, $x^2y$ which fill out $G$.

Where $x$ is of order $3$ and $y$ is of order 2. $x$, $y$ are both elements of $G$.

My confusion:

How can one guarantee that the right coset $\langle x\rangle$ and $\langle x\rangle y$ can fill out $G$? Because $\langle x\rangle$ and $\langle x\rangle y$ have no intersections? I am not sure about the properties of cosets. I also don't know if this is related to the fact that $|\langle x\rangle|=5$ and is precisely half the elements of $G$ and thus its left coset $y\langle x\rangle$ and right coset $\langle x\rangle y$ is exactly the same?

I don't know if you can understand my question. The question here actually arises from my vague understanding of "cosets". So everytime this term occurs I feel a steady uncertainty.

  • 1
    $\begingroup$ You may want to reread the question yourself and see where one or more letters can be added. $\endgroup$ – user21436 Mar 28 '12 at 1:36
  • $\begingroup$ @KannappanSampath: An artifact of using < and > in text; parser thought they were mark-up. $\endgroup$ – Arturo Magidin Mar 28 '12 at 2:10
  • $\begingroup$ This is specifically page 70 (and end of 69). $\endgroup$ – anon Mar 28 '12 at 2:12
  • $\begingroup$ Exactly. It is from page 70. Sorry for the disorganized order, I will check that next time before I submit. And thanks again to @ArturoMagidin for the edition. $\endgroup$ – Jinji Mar 28 '12 at 2:26
  • $\begingroup$ For more on cosets, see this answer. $\endgroup$ – Arturo Magidin Mar 28 '12 at 2:45

Let $H=\{e,x,x^2\}$ be the subgroup generated by $x$. The right cosets of $H$ are the sets of the form $Hz = \{hz\mid h\in G\}$ for fixed $z\in G$.

The right cosets of $H$ are known to form a partition of the set $G$; that is: any two cosets are either identical or disjoint, and their union is all of $G$.

Moreover, $Hz = Hw$ if and only if $zw^{-1}\in H$. This can be verified because $z\in Hz$ (obtained as $ez$), so there must exist some $h\in H$ such that $z = hw$. Hence $zw^{-1}=h\in H$, proving that if $Hz=Hw$, then $zw^{-1}\in H$. Conversely, if $zw^{-1}\in H$, then $z = (zw^{-1})w \in Hw$, so $Hz\cap Hw\neq\varnothing$; since right cosets are either disjoint or identical, and $Hz$ and $Hw$ are not disjoint, then they are identical.

Now, since $y$ is of order $2$, it is not in $H$ (every element of $H$ is either of order $3$ or of order $1$). That means that the cosets $He = H$ and $Hy=\{ey, xy, x^2y\}$ are distinct, hence they are disjoint. Since they are disjoint, $H$ and $Hy$, together, account for $6$ elements of $G$. Since $G$ has exactly $6$ elements by assumption, those are all the elements of $G$. That is: $$G = H\cup Hy = \{e,x,x^2\}\cup\{y,xy, x^2y\} = \{e,x,x^2,y,xy,x^2y\}.$$

The same is true for left cosets: left cosets of $H$ partition $G$, so any two left cosets are either disjoint or identical. Also, $zH = wH$ if and only if $z^{-1}w\in H$ (I'll leave the proof to you). That means that the two left cosets of $H$ are $eH=H$ and $yH$, since $yH\neq H$, so $yH\cap H=\varnothing$. Since $G$ has $6$ elements, $H$ has $3$, and $yH$ has another three, then $$G = H\cup yH = \{e,x,x^2\} \cup \{y, yx, yx^2\}.$$ Also, since $H\cap Hy=\varnothing$, then $Hy = G\setminus H$; similarly, $yH=G\setminus H$. So in fact, $Hy=yH$ as sets.

  • $\begingroup$ Thanks! Does the number of right cosets relate to the order of H? |G|/|H|=the number of right cosets? $\endgroup$ – Jinji Mar 28 '12 at 2:24
  • $\begingroup$ @Jinji: The number of cosets is, by definition, the index of $H$ in $G$, denoted $[G:H]$. Since the cosets all have the same size, namely $H$, and partition $G$, it follows that $|G|=|H|[G:H]$ (in the sense of cardinalities); in fact, if $K\lt H\lt G$, then $[G:K]=[G:H][H:K]$ in the sense of cardinalities. So in the special case in which $|G|$ is finite, it follows from $|G|=|H|[G:H]$ that $[G:H]=|G|/|H|$. $\endgroup$ – Arturo Magidin Mar 28 '12 at 2:37

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