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I am a newbie to group theory. I am confused by something that I would like clarified.

Say $G$ is the integers, a group under addition, and $H$ is the even integers, a subgroup of $G$.

It is usually stated that left cosets of $G$ are a partition of $G$.

Now for all $g \in G$, the left coset of $G$ by $g$ is defined as $gH$ in multiplicative notation, which for this example is really $g+H$.

So, here a left coset of $H$ by $g$ should be the set $\left\{g+h_1, g+h_2, g+h_3,\ldots\right\}$, where $h_1, h_2,h_3,\ldots$ are the elements of $H$.

How can these sets be disjoint? For example, the two cosets $2+H$ and $4+H$ both contain the element 8.

I don't understand what am I getting it wrong.

Thanks in advance.

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  • $\begingroup$ Only different cosets are disjoint, $2+H$ and $4+H$ are the same coset. $\endgroup$ – Hagen von Eitzen Nov 26 '13 at 21:16
  • $\begingroup$ Can you please elaborate ? My reading material defines coset as "Let H be a subgroup of a group G. The left coset of G relative to H are defined by g.H = {g.h|h ∈ H}, ∀g ∈ G." How are 2+H and 4+H 'same' cosets if they are different elements in G? $\endgroup$ – Koustav Ghosal Nov 26 '13 at 21:21
  • $\begingroup$ $2+2\mathbb{Z}=4+2\mathbb{Z}=2\mathbb{Z}=H$, because you said that $H$ is the set of even integers. $\endgroup$ – Dietrich Burde Nov 26 '13 at 21:24
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The distinct cosets of $H$ partition $G$. If $g_1+H\cap g_2+H \neq \emptyset$, then $g_1 + 2m_1 = g_2 + 2m_2$ for integers $m_1$ and $m_2$, and therefore $g_2 = g_1 + 2n$ for $n = m_2-m_1$. It follows that $g_1 + H = g_2 + H$.

The more precise statement is that if we choose a complete family of representatives $g_1,\ldots,g_n$ (or more generally $g_i$ for $i \in I$ with $I$ a (possibly infinite) index set), then $G$ is the disjoint union of $g_i + H$ for $i = 1,\ldots,n$.

Here a complete family of coset representatives is a set of distinct elements $g_1, \ldots, g_n$ of $G$ such that the cosets of $H$ in $G$ are exactly $g_1+H, \ldots, g_n+H$ and $g_i + H \neq g_j + H$ for $i \neq j$.


Edit: For your particular example, notice that $2+H = 4+H$. Indeed $g \in 2+H$ if and only if $g = 2+2n$ for some integer $n$, which in turn is true if and only if $g = 4 + 2m$ for some integer $m$. (Here $m = n-1$ since $g = 2+2n = 4 + 2(n-1)$.) Consequently $g \in 2+H$ if and only if $g \in 4+H$, i.e., $2+H = 4+H$.

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  • $\begingroup$ Okay,I think I am getting it. can you please suggest examples of two disjoint cosets in my case ? $\endgroup$ – Koustav Ghosal Nov 26 '13 at 21:38
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    $\begingroup$ @KoustavGhosal: Sure! In fact, in your case there are only two disjoint cosets. You can always choose $0 + H = H$ as one coset. Notice that $g+H \neq H$ if and only if $g$ isn't an even integer. So $1 + H$ is another coset, and the statement "$H$, $1+H$ partition $G$" is another way to say that "an integer is either even or odd (and never both)". $\endgroup$ – Dan Nov 26 '13 at 21:51

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