I have these notations in an exercise and I can’t understand them, the exercise is in French and I tried to translate it to English.

(“e” is the neutral element, “*” is a law of composition) (?)

1) Let (G,*) a group such as:

âˆ€x âˆˆ G, xآ²= e

-Is xآ² <=> x*x ?

2) Let (E,*) a group such as:

âˆ€x âˆˆ E, x*آ² = e

-What does x*آ² mean?

Thank you.

Original material(Ex 1 & 2): http://mp.cpgedupuydelome.fr/pdf/Structures%20alg%C3%A9briques%20-%20Groupes.pdf

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Could you provide a reference for the original material please?

– Larry B.

Oct 20 at 20:10

here, exercise 1 and 2: mp.cpgedupuydelome.fr/pdf/…

– مپ†مپ،م‚ڈ ه¯†مپ‹

Oct 20 at 20:12

1) Yes, x2=xxx^2=xx. 2) Does not apply for x∈Gx\in G, since EE is the group. This is different in the original.

– Dietrich Burde

Oct 20 at 20:15

I’m sorry it was a mistake, it’s xâˆˆE, I fixed it.

– مپ†مپ،م‚ڈ ه¯†مپ‹

Oct 20 at 20:18

Instead of an equivalence sign <=> between x2x^2 and x∗2x*2, use an equal sign, rigueur oblige.

– JeanMarie

Oct 20 at 21:08

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2 Answers

2

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Look very carefully at the original wording:

Soit (G,∗)(G, *) un groupe tel que…

So that GG is a group. However, in the second exercise:

Soit ∗* une loi de composition interne sur un ensemble EE associative et possأ©dant un neutre ee. On suppose que…

The key word here is “ensemble”, which means “set”. That is, EE is not defined as a group, let alone an abelian group. You must prove that EE is a group on ∗*.

In the group GG, the operator is bundled as part of the group, and x2x^2 is unambiguous (x2=x∗xx^2 = x * x). However, for the case of the set EE, the term x2x^2 is totally ambiguous. You are told that you have an associative operator ∗* on the set EE, not that EE is a group with the operator ∗*. Therefore we require the operator ∗* in the power to distinguish, “You are performing the binary operation ∗* upon the element xx two times.”

EDIT: You may be interested in this French glossary of math terms if you wish to do this translation. This is an elementary introduction to abstract algebra. Unless you have a reason to prefer this specific set of exercises, you may find it easier to learn from a text written in a language you’re fluent with. This introductory material is standard for any course in abstract algebra.

True, I’m a first year college student and these things are new to me even though they are the basics of abstract algebra,

– مپ†مپ،م‚ڈ ه¯†مپ‹

Oct 20 at 20:39

So, in the group G, x² is the same as x*x? For E, I understand it now.

– مپ†مپ،م‚ڈ ه¯†مپ‹

Oct 20 at 20:40

Yes. I’ll add that to the answer.

– Larry B.

Oct 20 at 22:34

You seem to have difficulties with groups. A group is a set GG together with an operation ∗*, that combines two elements in GG to form a new element in GG. The group is mostly written like this: (G,∗)(G,*). A group must satisfy certain laws:

Closure: When you combine two elements in GG with the operation

∗*, the third element also has to be part of GG,

Assosiativity: Take 3 elements in GG: (a,b,c)(a,b,c), then the following

must apply: (a∗b)∗c=a∗(b∗c)(a*b)*c=a*(b*c)

Neutral element: There exists an element ee in GG, so that for all

xx in GG, x∗e=xx*e = x

Inverse: For all elements xx in GG, there exists an element x′x’

in GG so that x∗x′=ex*x’=e (ee being the neutral element)

An example of a group would be (Z,+)(\mathbb Z,+), the neutral element being 00, and the inverse beingthe negative value of a given number in Z\mathbb Z.

Note that the operation ∗* is not restricted to +,∗+,*, as any operation that satisfies the group definition works.

Referring to your question, it should now be clear as to what ee and ∗* refer, x2x^2 is indeed x∗xx*x. As for x∗xx^{*x}, I can not help you, as I do not see what it is supposed to mean. If you have problems with logical quantifiers, refer to

https://en.wikipedia.org/wiki/Quantifier_(logic).

Thank you, much appreciated.

– مپ†مپ،م‚ڈ ه¯†مپ‹

Oct 20 at 21:20