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I.N. Herstein has defined:

Let $G$ be a group, $H$ a subgroup of $G$; for $a,b \in G$ we say $a$ is congruent to $b \mod H$, written as $a \equiv b \mod H$ if $ab^{-1} \in H$.

Let $G$ be a group, $H$ a subgroup of $G$; for $a,b \in G$ we say $a$ is congruent to $b \mod H$, written as $a \equiv b \mod H$ if $ab^{-1} \in H$.

When dealing with integers $a \equiv b \mod n$ means $n \mid (a-b)$, what is the meaning of $a \equiv b \mod H$ in context of group theory? And why and how has $ab^{-1} \in H$ been used as a condition to define this relation?

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When dealing with the integers, our subgroup $H$ is the group $n\mathbb{Z}$ and must recall that the operation on the integers in this case is additive, so $b^{-1}$ is actually $-b$. Thus, if $a \equiv b\pmod{n}$, then $a-b \in n\mathbb{Z}$, so $a - b$ is a multiple of $n$.

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  • $\begingroup$ But he hasn't specified that the operation is additive in nature or that is implied? $\endgroup$ – user88923 Jun 27 '14 at 4:42
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    $\begingroup$ When regarding the integers as a group, you must take the operation to be addition since you don't have multiplicative inverses (other than $\pm 1$). $\endgroup$ – Jacob Bond Jun 27 '14 at 4:44
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Remember that a group has only one operation, and as far as the notation goes you can call it anything you like. It might help you to understand Herstein's definition if we rewrite it using additive notation instead of multiplicative. Thus we write $+$ instead of $\times$, and we write $-a$ instead of $a^{-1}$. So, Herstein rewritten:

Let $G$ be a group, $H$ a subgroup of $G$; for $a,b\in G$ we say $a$ is congruent to $b$ mod $H$, written as $a\equiv b\ {\rm mod}\ H$, if $a-b\in H$.

I hope this makes the analogy with integer congruences clear.

To make it even a bit more than an analogy, consider the case when the group $G$ is the set of integers, the operation is addition (in the usual sense), and the subgroup $H$ is $n\Bbb Z$, the set of all multiples of $n$. Then $a\equiv b\ {\rm mod}\ H$ means that $a-b$ is in $n\Bbb Z$, that is, $a-b$ is a multiple of $n$, that is, $n\mid a-b$ just as in the usual definition of congruence.

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  • $\begingroup$ But he hasn't specified that the operation is additive in nature or that is implied? $\endgroup$ – user88923 Jun 27 '14 at 4:40
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    $\begingroup$ No, the point is that it doesn't matter what the operation is. There is only one operation. You can call it multiplication, that doesn't mean it really is multiplication; you can call it addition, that doesn't mean it really is addition. Writing the definition in terms of addition does not change the meaning, it only (hopefully) makes it easier to understand the correspondence with normal congruence of integers. $\endgroup$ – David Jun 27 '14 at 4:46
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The $a \equiv b \pmod{H}$ is a generalization of the "mod" we see in integers. Recall that $(\mathbb{Z},+)$ is a group. So $ab^{-1}$ is the generalization of $a-b=a+(-b)$.

Now comes the other part: $a \equiv b \pmod{n}$ means $n|a-b$. This can be interpreted as $a-b \in n\mathbb{Z}$, where $n\mathbb{Z}=\{nk \, | \, k \in \mathbb{Z}\}$. Observe that all subgroups of $\mathbb{Z}$ are of the form $n\mathbb{Z}$. So to generalize this for any group $G$ and any subgroup $H$. We interpret $a \equiv b \pmod{H}$ as $ab^{-1} \in H$.

I hope this helps.

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If $n | (a - b)$, then $(a - b) = kn, \exists k \in \mathbb Z$ or equivalently, $a - b \in n \mathbb Z$ (where $n \mathbb Z \equiv \{kn \mid k \in \mathbb Z\}$ is a subgroup of the group $\mathbb Z$ under addition). We can also write this as $ab^{-1} \in H$, where $b^{-1} = -b$ is the inverse of $b$ in the group (the additive inverse), and $H = n\mathbb Z$. Also note that every subgroup of $\mathbb Z$ is of the form $n\mathbb Z$.

The same can be generalized to a general group and its subgroups.

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  • $\begingroup$ But he hasn't specified that the operation is additive in nature or that is implied? $\endgroup$ – user88923 Jun 27 '14 at 4:43
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    $\begingroup$ Just to add to the answer: like it was said, if $a \equiv b \mod n$ then $a - b = kn $ for some $k \in \Bbb Z$. For groups, $a \equiv b \mod H$ means that $ab^{-1} \in H $, so $ab^{-1} = h$, hence $a = hb$ for some $h \in H$. $\endgroup$ – Ivo Terek Jun 27 '14 at 4:44
  • $\begingroup$ @MadRumours I was specifically talking about the group of integers, and observing that $a - b$ is nothing but $ab^{-1}$ written in additive notation. $\endgroup$ – M. Vinay Jun 27 '14 at 4:52

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