# Algebraic substructure and restriction of a function

I am reading the follow pdf:

http://www.math.uwaterloo.ca/~snburris/htdocs/UALG/univ-algebra2012.pdf

in particular at pg 28 of pdf, and I think that, let $(A;f)$ an algebric structure and $B \subseteq A$ and $B \neq \emptyset$,$B$ is an algebraic substructure of $(A;f)$ if $(B;f_{|B})$ is an algebric structure and $f_{|B}$ is restriction function on $B$. It is correct?

Thanks in advance!!

-

## 1 Answer

Yes, basically correct.

An algebraic structure may have more operations (e.g. ring).

The main thing for $B$ being subalgebra of $A$ is, that, each operation $f:A^n\to A$ stays within $B$ when applied to elements of $B$ (also said as "$B$ is closed under the operation $f$"), that is: $f|_{B^n}:B^n\to A$ factors through $B$: $f(b_1,..,b_n)\in B$ for all $b_i\in B$.

In notation the restriction to $B^n$ of an $n$-ary operation $f$ is frequently written briefly $f|_B$, i.e. technically $f|_B$ can denote the restriction of $f$ to $B^n\subseteq A^n$.

-
but can I use this definition: proofwiki.org/wiki/Definition:Restriction_of_Mapping#Definition?? In this case I have the restriction of $f$ to $(B \times B) \times B$, or in simbol $f{\restriction_{(B \times B) \times B}}$, with $f(B) \subseteq B$ or in other words $B$ has the property of closure under $f$... Is correct?? Thanks in advance!! – mle Apr 21 '13 at 16:13