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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 finite

Let G be a group. If G is finite there is only a finite number of subsets of G, so clearly

a finite number of subgroups.

Now suppose G is infinite , let's

suppose one element has infinite order. The this element generates an infinite cyclic

group 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.

Q: G is finite where the number of subgroups in g is finite?

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Let H and I be subgroups of G. A group B is a subgroup of a group A if and only if every member of B is a member of A. Thus, every member of H is a member of G, and every member of I is a member of G. The intersection of two groups A and B is the set of things that are members of both A and B. Hence, if a group C is equivalent to the intersection of A and B, then everything that is a member of both A and B is a member of C, and everything that is a member of C is a member of both A and B. Hence, everything that is a member of the intersection of H and I is a member of H. We established above that every member of H is a member of G. Thus, everything that is a member of the intersection of H and I is a member of G. Recall that A group B is a subgroup of a group A if and only if every member of B is a member of A. Hence, if everything that is a member of the intersection of H and I is a member of G, then the intersection of H and I is a subgroup of G, and so the intersection of H and I is indeed a subgroup of G.

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Related questions

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

(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.

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.

The cardinality of a set is its size. For instance, since the set G contains 4 elements, then its cardinality is 4. So if the set has a finite number of elements (meaning it is a finite set), you can find its cardinality, otherwise you cannot (meaning it is an infinite set).

This follows immediately from the first Sylow theorem.

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