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False. G may be a finite group without sub-groups.

Q: G is finite group if and only if the number of it is subgroups is finite?

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There is only a finite number of opportunities in your life.

Lagrange theorem states that the order of any subgroup of a group G must divide order of the group G. If order p of the group G is prime the only divisors are 1 and p, therefore the only subgroups of G are {e} and G itself. Take any a not equal e. Then the set of all integer powers of a is by definition a cyclic subgroup of G, but the only subgroup of G with more then 1 element is G itself, therefore G is cyclic. QED.

(1). G is is finite implies o(G) is finite.Let G be a finite group of order n and let e be the identity element in G. Then the elements of G may be written as e, g1, g2, ... gn-1. We prove that the order of each element is finite, thereby proving that G is finite implies that each element in G has finite order. Let gkbe an element in G which does not have a finite order. Since (gk)r is in G for each value of r = 0, 1, 2, ... then we conclude that we may find p, q positive integers such that (gk)p = (gk)q . Without loss of generality we may assume that p> q. Hence(gk)p-q = e. Thus p - q is the order of gk in G and is finite.(2). o(G) is finite implies G is finite.This follows from the definition of order of a group, that is, the order of a group is the number of members which the underlying set contains. In defining the order we are hence assuming that G is finite. Otherwise we cannot speak about quantity.Hope that this helps.

Rational

It is a requirement to find a decimal representation which has only a finite number of digits after the decimal point.

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Actually a stronger statement can be made:A group G is finite if and only if the number of its subgroups is finiteLet G be a group. If G is finite there is only a finite number of subsets of G, so clearlya finite number of subgroups.Now suppose G is infinite , let'ssuppose one element has infinite order. The this element generates an infinite cyclicgroup which in turn contains infinitely many subgroups.Now suppose all the subgroups have finite order Take some element of G and let it generate a finite group H. Now take another element of G not in H and let it generate a finite group I. Keep doing this by next picking an element of G not H or I. You can continue this way.

There is only a finite number of opportunities in your life.

It was presumably proven when it was discovered that there were infinitely many counting numbers. However, whoever it was, did not consider the mathematical possibility with practicality. The universe has a finite life. Within that our solar system is finite. People, in their turn, have finite lives. In a finite life you can only "dial" a finite number of digits. therefore, you can only call a number if it has a finite number of digits. For any finite number of digits, there are only a finite amount of phone numbers. So, having infinitely many telephone numbers is no use if you need to wait an infinite amount of time (longer than you'll live) for the first person to call you!

Lagrange theorem states that the order of any subgroup of a group G must divide order of the group G. If order p of the group G is prime the only divisors are 1 and p, therefore the only subgroups of G are {e} and G itself. Take any a not equal e. Then the set of all integer powers of a is by definition a cyclic subgroup of G, but the only subgroup of G with more then 1 element is G itself, therefore G is cyclic. QED.

Every human being born on this planet has a soul. So the number always remains a finite number. So the belief is true.

The defining characteristic of FA is that they have only a finite number of states. Hence, a finite automata can only "count" (that is, maintain a counter, where different states correspond to different values of the counter) a finite number of input scenarios.There is no finite automaton that recognizes these strings:The set of binary strings consisting of an equal number of 1's and 0'sThe set of strings over '(' and ')' that have "balanced" parenthesesThe 'pumping lemma' can be used to prove that no such FA exists for these examples.

(1). G is is finite implies o(G) is finite.Let G be a finite group of order n and let e be the identity element in G. Then the elements of G may be written as e, g1, g2, ... gn-1. We prove that the order of each element is finite, thereby proving that G is finite implies that each element in G has finite order. Let gkbe an element in G which does not have a finite order. Since (gk)r is in G for each value of r = 0, 1, 2, ... then we conclude that we may find p, q positive integers such that (gk)p = (gk)q . Without loss of generality we may assume that p> q. Hence(gk)p-q = e. Thus p - q is the order of gk in G and is finite.(2). o(G) is finite implies G is finite.This follows from the definition of order of a group, that is, the order of a group is the number of members which the underlying set contains. In defining the order we are hence assuming that G is finite. Otherwise we cannot speak about quantity.Hope that this helps.

It has 4 subgroups isomorphic to S3. If you hold each of the 4 elements fixed and permute the remaining three, you get each of the 4 subgroups isomorphic to S3.

Finite VerbsA finite verb (sometimes called main verbs) is a verb that has a subject, this means that it can be the main verb in a sentence. It shows tense (past / present etc) or number (singular / plural).For example:-I live in Germay. (I is the subject - livedescribes what the subject does - live is a finite verb).Non-Finite VerbsA non-finite verb has no subject, tense or number. The only non-finite verb forms are the infinitive (indicated by to), the gerund or the participle.For example:-I lived in Germany to improve my German. (To improve is in the infinitive form - improve is non-finite).

Income is discrete. People can have only a finite number of possible income values.

It can take only a finite number of values. These need not be integer values.

The money multiplier is the reciprocal of the reserve requirement, which can only be a finite number.