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The Lemma asserted in Herstein as given by $[a] = Ha$ seems very non intuitive to me. How do I think in order that this thing makes sense to me?

LEMMA 2.4.4 For all $a$ in $G$ , $$Ha = \{ x \in G : a \equiv x \mod H\}$$

Thanks

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  • $\begingroup$ The way I do it is purely algebraically. Just try it as the set with the properties provided and try to manipulate them. Work out some examples for yourself (like the Klein 4 group, or $\mathbb{Z}$ and $n\mathbb{Z}$ for some $n$'s, or $S_3$) and just write out the sets themselves (and the set of sets). $\endgroup$ – Eoin Mar 23 '15 at 7:13
  • $\begingroup$ Imagine lining up the elements of $H$ in a vertical column. At the top is the identity $e$. Multiplying by $a$ shifts everything to the right, so $Ha$ is a column of elements with the topmost element $a$. This illustrates that all of the cosets have the same cardinality and partition the group. $\endgroup$ – William Stagner Mar 23 '15 at 7:14
  • $\begingroup$ @K.Dutta Please give the statement of the Lemma in the question. Otherwise I guarantee you this question is on a short trip to "on hold: missing context". $\endgroup$ – Mario Carneiro Mar 23 '15 at 7:14
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    $\begingroup$ For any equivalence relation, represent elements of your original set as mushrooms (or favorite topping). Place them on pizza and slice it up. All mushrooms on a slice are in one partition/coset. $\endgroup$ – dls Mar 23 '15 at 15:13
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It's easiest to see this if we start with the group of integers under addition (in this operation $Ha$ is wriiten $H + a$). Imagine the subgroup of multiples of $3$:

$3\Bbb Z = \{\dots,-9,-6,-3,0,3,6,9,12,\dots\}$

Then $3\Bbb Z + 1$ is $3\Bbb Z$ "shifted by $1$" (just add one to every element of $3\Bbb Z$):

$3\Bbb Z + 1 = \{\dots,-8,-5,-2,1,4,7,10,13,\dots\}$, and $3\Bbb Z + 2$ works the same way:

$3\Bbb Z + 2 = \{\dots,-7,-4,-1,2,5,8,11,14,\dots\}$.

Note that $3\Bbb Z + 3$ is just $3\Bbb Z$ again, since $3\Bbb Z$ has "gaps" of $3$, and extends infinitely negatively and positively. So we only get those $3$ cosets, which chops the integers into $3$ pieces, integers of the form $3k,3k+1$, or $3k+2$, and each "piece" of the integers is exactly the same "size".

So, now let's look at some finite group. Suppose we have a subgroup $H = \{e,h_1,\dots,h_n\}$. To get $Ha$ we "multiply everything by $a$ (on the right)":

$Ha = \{a,h_1a,\dots,h_na\}$.

We can do this for ANY $a \in G$, so it is natural to ask: can we have $Ha = Hb$?

If so, then for any $h_ka \in Ha$, it must be equal to some $h_mb \in Hb$. This means:

$h_ka = h_mb \implies h_k = h_mba^{-1} \implies h_m^{-1}h_k = ba^{-1}$.

So if $Ha = Hb$, then $ba^{-1} \in H$. So that is what we call a "necessary" condition for $Ha$ to equal $Hb$. When encountering a necessary condition, mathematicians like to inquire if it is "sufficient" as well. So we start with the OTHER assumption:

$ba^{-1} \in H$. What will this tell us about $Ha$ and $Hb$? Well, if $ba^{-1} = h \in H$, then $ha = b$.

Note we can re-write $h_k \in H$ as: $h_kh^{-1}h$. So:

$Ha = \{a,h_1a,\dots,h_na\} = \{h^{-1}ha, h_1h^{-1}ha,\dots,h_nh^{-1}ha\}$

$= \{h^{-1}b, h_1h^{-1}b,\dots,h_nh^{-1}b\}$

What we'd like to show now, is that the mapping $h_k \mapsto h_kh^{-1}$ ($H \to H$) is bijective.

So suppose $h_kh^{-1} = h_mh^{-1}$. Multiplying both sides on the right by $h$, gives $h_k = h_m$, so our mapping is injective. On the othr hand, given any $h_j \in H$, we see that: $h_jh \in H$, and our mapping takes:

$h_jh \mapsto (h_jh)h^{-1} = h_j(hh^{-1}) = h_j$, so the map $h_k \mapsto h_kh^{-1}$ is bijective.

THIS tells us the set $\{h^{-1}b, h_1h^{-1}b,\dots,h_nh^{-1}b\} = Hb$ since $h_kh^{-1}$ runs through ALL elements of $H$, just like $h_k$ does.

In other words $Ha = Hb$ if, and only if, $ba^{-1} \in H$ (since $H$ is a subgroup, this also means $(ba^{-1})^{-1} = ab^{-1} \in H$, as well).

So we now define a relation: $a \sim_H b$ means: $Ha = Hb$. As we saw above this means that $ba^{-1}$ (and $ab^{-1}$) are in $H$. It is not immediately apparent this is an equivalence relation. We have to PROVE it.

$\sim_H$ is reflexive: this states that $Ha = Ha$, or equivalently, that $aa^{-1} = e \in H$. This is obvious, since $H$ is a subgroup, and subgroups contain the identity.

$\sim_H$ is symmetric: this says if $Ha = Hb$, then $Hb = Ha$. Well that seems clear. Alternatively, if $ba^{-1} \in H$, then since $H$ is a subgroup, $(ba^{-1})^{-1} = ab^{-1} \in H$, that is: $a\sim_H b \iff b \sim_H a$.

$\sim_H$ is transitive: this says if $Ha = Hb$, and $hb = Hc$, then $Ha = Hc$. Again, this seems straight-forward. Using our "other definition" we need to show that if:

$ba^{-1}$ and $cb^{-1}$ are in $H$, so is $ca^{-1}$, But $H$ is a subgroup, and so is closed under multiplication, so:

$(cb^{-1})(ba^{-1}) = c(b^{-1}b)a^{-1} = ca^{-1} \in H$.

This tells us $\sim_H$ is a bona-fide equivalence relation. It should be clear now that if $b \in [a]_{\sim_H}$, that $Hb = Ha$, which means in particular, $b = eb \in Ha$. On the other hand, if $b \in Ha$, so that $b = ha$ for some $h \in H$, then $ba^{-1} = h \in H$, so that $b \in [a]_{\sim_H}$.

In summary, the equivalence class of $a$ (under the equivalence relation $\sim_H$), $[a]$ is PRECISELY the (right) coset, $Ha$.

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  • $\begingroup$ Can you explain why if Ha=Hb implies $ba^{-1}$ in H . How does $(h_{m})^{-1}h_{k} = ba^{-1}$ proves it ? $\endgroup$ – Taylor Ted Mar 23 '15 at 9:02
  • $\begingroup$ @K.Dutta Because $H$ is a subgroup, it is closed under inverse and group operation, so $h_m\in H\implies (h_m)^{-1}\in H\implies (h_m)^{-1}h_k\in H$. $\endgroup$ – Mario Carneiro Mar 23 '15 at 10:03
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The coset of a group I like to think of it as the set H multiplied by element a if your group is normal that is Ha = aH then multiplication by a will make both sets equal.

I will try to explain to you the whole story we have in normal $Z_n$ we have the following a $\equiv$ b (mod n) iff a - b = n q1*

we could look at this from another perspective that is we could instead of looking as numbers we could look at this in terms of sets that is we would have

a $\equiv$ b (mod Z_n) iff $a*b^{-1} \in Z_n$

we can show that this relation gives an equivalence relation and hence congruence class partitions your group G and the classes are given as follows

{b $\in$ G | b $\equiv$ a (mod K)} = {b $\in$ G | $b*a^-1$ = k, such that k $\in K$} = {ka | k $\in K$}. So the congruence class of a modulo K is our right coset.

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It seems that the first $[a]$ is the equivalence class formed by all elements equivalent to $a$ under the relation $a\sim b$ when there is an $h \in H$ so that $a=hb$. While $Ha$ is the set formed by all products of $h \in H$ with a. These are the same because if $a = hb$ then $b = h^{-1}a$ and $h^{-1}\in H$ iff $h \in H$, so that the collection of all such elements $b$ is the same as that of all products in $Ha$. Intuitively one can think of a coset as some kind of orbit formed by the action of $H$ on an element $a \in G$, for example if $a$ is an element of $H$ then the orbit is $H$ itself, for $H$ is closed under multiplication (being a subgroup).

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  • $\begingroup$ What do you mean by orbit ? $\endgroup$ – Taylor Ted Mar 23 '15 at 7:19

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