What exactly is an "induced operation"?

Question

Examples of places where I see this used:

Let * be a binary operation in S and let H be a subset of S... the binary operation on H given by restricting * to H is the induced operation of * on H

For a subgroup, it is not sufficient that "the set of one group be a subset of another, but also that the group operation on the subset be the induced operation that assigns the same element to each ordered pair from this subset as is assigned by the group operation on the whole set"

I just really don't understand what an induced operation is supposed to be.

Is this an accurate understanding?: If you have the binary structure $\langle S,* \rangle$, then the induced operation is if you make another binary structure with set S' and let the operation be the same * as in the first binary structure. In other words, is it accurate to say that the operation * in the binary structure $\langle S',* \rangle$ is the induced operation of $\langle S,* \rangle$, but the operation *' in $\langle S',*' \rangle$ is not?

All quotes from the book "A First Course in Abstract Algebra" by John Fraleigh, 7th edition

Answer 1

Your suggestion is correct, except that you forgot to say what $S'$ has to do with $S$. And it is this: $S'$ is a subset of $S$. For instance, sum on $\mathbb R$ does not induce an operation in $(-1,1)$ since, in general the sum of two elements of $(-1,1)$ is not an element of $(-1,1)$.

There is another case of induced operations, which appears in the context of quotients of algebraic structures, but I don't know whether or not you are familar with this.

Answer 2

I think looking at an example is quite telling: let us consider $\mathbb{Q}\subset \mathbb{R}$. Then $\mathbb{R}$ and $\mathbb{Q}$ are groups for the usual addition. But $\mathbb{Q}$ is also a group for the (quite arbitrary) operation $a\star b= a+b+1$.

Now the addition on $\mathbb{Q}$ is the operation induced by the addition on $\mathbb{R}$, meaning that if you take two elements $a,b\in \mathbb{Q}$, you can add them as real numbers or as rational numbers, and it is the same thing.

On the other hand, even though $(\mathbb{Q},\star)$ is a group and $\mathbb{Q}\subset \mathbb{R}$, the operation $\star$ is not the restriction of the addition on $\mathbb{R}$, it is not the induced operation, so this does not define a subgroup.

On the other hand, if we define $a\bullet b= a+ b+1$ on $\mathbb{R}$, then $\star$ on $\mathbb{Q}$ is the restriction of $\bullet$ on $\mathbb{R}$, so $(\mathbb{Q},\star)$ is a subgroup of $(\mathbb{R},\bullet)$.

Answer 3

Induced operation. Definition (2.4) Let $\ast$ be a binary operation on a set $S$ and let $H$ be a subset of $S$. If for all $a,b \in H$ we also have $a\ast b \in H$, then $H$ is closed under $\ast$. In this case, the binary operation on $H$ given by restricting $\ast$ to $H$ is the induced operation of $\ast$ on $H$.

The above is quoted from Section 2 – Binary Operations Instructor: Yifan Yang Fall 2006 the first result on google search for induced binary operation.

My understanding would be it induces a new magma ( a set with an operation it's closed under) which if it has certain elements would be a subgroup, subring, etc.