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first my definition of a coset that I got is: Definition. Let $H$ be a subgroup of a group $G$. A left coset of $H$ in $G$ is a subset of $G$ that is of the form $xH$, where $x\in G$ and $$xH = \{y\in G | y = xh \text{ for some }h\in H\}.$$

And here is the theorem that is confusing me.

Let $H$ be a subgroup of a group $G$. Then the left cosets of $H$ in $G$ have the following properties:

  1. $x\in xH$ for all $x\in G$;
  2. if $x$ and $y$ are elements of $G$, and if $y = xa$ for some $a\in H$, then $xH = yH$;
  3. if $x$ and $y$ are elements of $G$, and if $xH \cap yH$ is non-empty then $xH = yH$.

$\textbf{Proof.}$ Suppose $x\in G$. Then $x = x1$, but $1$ is in $G$ and $H$ since $H$ is a subgroup of $G$. Thus $x\in xH$, which proves (1).

Suppose $x$ and $y$ are elements of $G$ and we can choose some $a\in H$ such that $y = xa$. Now suppose $z\in yH$. It follows that $z = yh$ for some $h\in H$. But $y = xa$, so we know that $z = xah$. Since $H$ is a subgroup and $a\in H$ and $h\in H$ it follows that $ah \in H$, hence $z\in xH$. Since $z$ was an arbitrary element of $yH$ we can conclude that $yH\subseteq xH$. Now suppose $z\in xH$. It follows that $z = xh$ for some $h\in H$. But $x =ya^{-1}$, so $z = ya^{-1}h$. Since $H$ is a subgroup and $a\in H$ it follows that $a^{-1}$ is in $H$. Similarly, since $a^{-1}\in H$ and $h\in H$ it follows that $a^{-1}h \in H$, thus $z\in yH$. Since $z$ was an arbitrary element of $xH$ we can conclude that $xH \subseteq yH$. Therefore we can conclude that $xH = yH$, which proves (2).

Now suppose that $x$ and $y$ are elements of $G$ and $xH \cap yH$ is non-empty. Since $xH \cap yH$ is non-empty we can choose some $z\in xH \cap yH$. Since $z\in xH\cap yH$ it follows that $z\in xH$ and $z\in yH$. Thus $z = xh_x$ with $h_x \in H$ and $z = yh_y$ with $h_y \in H$. Thus, since $z = z$ in all groups, $yh_y = xh_x$. Hence, $y =xh_xh_y^{-1}$. Since $H$ is a subgroup $x\in H$, $h_x\in H$ and $h_y\in H$, we can conclude that $xh_xh_y^{-1} \in H$. Therefore by (2) we know that $xH = yH$, which proves (3). $\Box$

Now my confusion stems from (1) and (3).

Suppose $G = \{x_1, x_2, ..., x_n\}$ under $\star$ Then each element $x_i$ of $\{x_1, x_2, ..., x_n\}$ is a member of $x_iH$ by (1).

By (3) we know that no element $x_i$ can be a member of two sets $x_iH$ and $x_jH$.

But since each $x_iH$ is a subset of $\{x_1, x_2, x_3, ..., x_n\}$, so it would seem to me that each coset would have to contain only one element.

Where am I wrong in my thinking here? Thank you very much

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3 Answers 3

up vote 1 down vote accepted

An element $x_i$ can be a member of $x_iH$ and $x_jH$ if $x_iH$ and $x_jH$ are simply different names for (or descriptions of) the same set.

Think about the coset $1_GH = H$ for starters: for every $h \in H$, $hH = H$. All of the cosets $hH$ with $h \in H$ are the same coset, namely, $H$ itself. While this is most obvious in the case of $H$, the same thing is true of all other cosets of $H$: if $C$ is any coset of $H$, then $gH = C$ for every $g \in C$.

You might try to prove that for any $x,y \in G$, $xH = yH$ if and only if $x^{-1}y \in H$; all of the ingredients can be found in your proof of (1)-(3).

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Thanks very clear answer, and thanks for the suggestion I will try proving it later when I can get to my computer –  Deven Ware Aug 11 '11 at 0:13
    
Thanks for editing your response :) –  Deven Ware Aug 12 '11 at 17:48

It might be that $x_1H = x_2H$, even though $x_1 \neq x_2$. In particular, if $h$ is in $H$, then $hH = 1H = H$, even though $h$ need not be equal to $1$.

In general, if $H$ has $m$ elements, then $xH$ has $m$ elements too. The cosets divvy up $G$ into pieces of size $m$. Each piece has $m$ different names, $(xh) H = xH$ for all $m$ possible $h$s in $H$.

If $G$ is the group of points in the Cartesian plane (the operation being addition), and $H$ is a line through the origin (for instance the $x$-axis, the $y$-axis, or the line $y=x$ all form subgroups), then the cosets of $H$ are all the parallel lines to $H$. Each line can be named by any of the points on it.

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Thank you, cleared up my cofusion so simply very much appreciate it :) –  Deven Ware Aug 11 '11 at 0:01

(iii) only says that $x_i H$ and $x_j H$ are either the same, or completely disjoint. Thus if you pick any element $y$ from $xH$, then it will necessarily be the case that $xH = yH$. Note that this does not imply that $x=y$, only that $xH = yH$.

For instance consider the group $G = \{1,x,y,xy\}$ where $x^2 = 1$ and $y^2 = 1$ and consider the subgroup $H = \{1,x\}$.

The coset $1 H = \{1, x\}$ and the coset $xH = \{x, x^2\} = \{1,x\}$. Thus $1H = xH$.

The coset $yH = \{y, yx\}$ which is disjoint from $1H$.

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Thank you very much, I'm no longer confused :) –  Deven Ware Aug 11 '11 at 0:02

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