Vtyjkyuk

I consider the Lagrange theorem.
Let $G$ be a finite group and let $H subseteq G$ be a subgroup, then the order of $H$ divides the order of $G$.
I am interesting with the proof of this theorem. The proof is as follows

Let $C= a_1 H, a_2 H,ldots,a_t H$ be a collection of all distinct left cosets of $H$ in $G$, where $t$ is a positive integer.
Since $C$ is a partition of $G$ ,we have
$|G|=|a_1 H|+|a_2 H|+ldots+|a_t H|=|H|+|H|+….+|H|=t|H|$, hence the theorem is proved!

Now how can I prove that the converse of this theorem is not true? Is the converse not true generally ?

Here is the way I try, from $|G|=|a_1 H|+|a_2 H|+ldots+|a_t H| = |H|+|H|+ldots+|H|=t |H|$. If I try to write $|H|/|G|=1/t$, I see that $1/t$ is not an integer, so that the converse of Lagrange theorem is not true. Does this satisfy for the proof “converse of Lagrange's theorem is not true”? Please I beg your more help for this case, thanks

When you have an implication, "if $P$, then $Q$", the converse is the implication "if $Q$, then $P$."

Lagrange's Theorem says:

Let $G$ be finite with $|G|=n$. If $d$ is the order of a subgroup of $G$, then $d$ divides $n$.

The converse would therefore say:

Let $G$ be a finite group with $|G|=n$. If $d$ divides $n$, then $d$ is the order of a subgroup of $G$.

This is an implication. In order to show that an implication is false, you need to show that it is possible for the antecedent ("$d$ divides $n$") to be true and at the same time, the consequence ("$d$ is the order of a subgroup of $H$") to be false.

So in order to show that the converse of Lagrange's Theorem is not true, what you need to do is:

1. Exhibit a specific finite group $G$;


2. Exhibit a specific number $d$ that divides $G$; and


3. Prove that $G$ does not have any subgroups of order $d$.

In other words, you need to give an example that shows that the converse statement does not have to be true. (A "counterexample".)

What you did makes no sense, and does not prove anything. The "converse" has nothing to do with the reciprocal $|H|/|G|$, and is certainly not an assertion that the order of $G$ divides the order of $H$.

Now, just to help in your search: if $G$ is abelian, then it will satisfy the converse of Lagrange's Theorem, so try looking for nonabelian groups; if $|G|$ is a power of a prime, or the product of two distinct primes, then $G$ will satisfy the converse of Lagrange's Theorem as well, so avoid those. But try the very smallest size that is left once you throw out all of those orders.

In the positive direction, it is easiest to note that if $G$ is a finite cyclic group of order $n$ and $k$ divides $n$, then $G$ has a subgroup of order $k$. It is generated by $a^n/k$ if $a$ generates $G$. This can be extended to (finite) abelian groups because they are direct sums of cyclic groups.

Every supersolvable group also satisfies the converse of Lagrange's Theorem, and every group that satisfies the converse of Lagrange's Theorem is solvable. This was shown by Bray in "A note on CLT groups," 1968. Every subgroup and every quotient of a supersolvable group is supersolvable, hence satisfies the converse of Lagrange's Theorem.

As Arturo noted, standard results of group theory show that there can be no counterexample whose order is a power of a prime or a product of two distinct primes. But there are other numbers, like $20=2^2cdot 5$, for which every group with that number of elements satisfies the converse of Lagrange's Theorem. There have been several articles written both on the possible orders of groups that don't satisfy the converse as well as other properties of such groups. One example is a result of Struik in "Partial converses to Lagrange's theorem,", 1978. To quote part of the review by Humphreys (because I don't have the article):

Let $p$ and $q$ be distinct primes with $p$ not dividing $q-1$ and $q$ not dividing $p-1$. Let $e$ be the exponent of $p textmod,q$ and $f$ be the exponent of $q textmod,p$. Then $p^aq^b$ is a CLT number if and only if one of the following four conditions is satisfied: (i) $a<e$ and $b<f$; (ii) $e$ is odd, $a=2e-1$ and $b<f$; (iii) $f$ is odd, $b=2f-1$ and $a<e$; (iv) $e$ and $f$ are both odd, $a=2e-1$ and $b=2f-1$.

For example, if $p=3$ and $q=5$, then $e=4$ and $f=2$, so by part (i) we see that every group of order $45=3^2cdot 5$ or $135=3^3cdot 5$ satisfies the converse of Lagrange's Theorem. It is known (as seen for example in Curran's "Non-CLT groups of small order" where such groups are studied) that the numbers less than $100$ that are orders of groups that don't satisfy the converse of Lagrange's Theorem are the following: $12,24,36,48,56,60,72,75,80,84$ and $96$.

(I didn't know any of the results from these articles, but I searched because the question made me interested.)

What the converse of a theorem is depends on exactly how it is formulated, and in the formulation you gave there is no converse. As Arturo said you need a statement of the form "if $P$ then $Q$" to turn into the converse "if $Q$ then $P$" (or if you prefer "if not $P$ then not $Q$, which is logically equivalent). However your statement does not contain the word "if", so strictly speaking there is no converse.

Arturo gave an equivalent statement that does contain the word "if", and proved that the converse to that statement is false. However his formulation is rather distant from yours, notably he introduces $d$ that is not mentioned in your statement at all (and the mention of $H$ is dropped). A closer approximation would be the formulation "Let $G$ be a finite group. If $H$ is a subgroup of $G$, then the order of $H$ divides the order of $G$". The converse of this statement is "Let $G$ be a finite group. If the order of $H$ divides the order of $G$, then $H$ is a subgroup of $G$". The problem with this statement is that $H$ now comes out of the blue; since we have to prove that $H$ is a subgroup of $G$, we may not assume that in interpreting the "if" part, and we have the problem that, without any assumption on $H$, the phrase "the order of $H$" makes no sense. So with this approach there is still no sensible converse to prove. A solution would be to expand "the order of $H$" to "the number of elements of $H$", which makes sense whenever $H$ is a set. If we add a clause stipulating that before the "if", we get as original statement "Let $G$ be a finite group and $H$ a set. If $H$ is a subgroup of $G$, then the number of elements of $H$ divides the order of $G$", which has a sensible (but clearly false) converse "Let $G$ be a finite group and $H$ a set. If the number of elements of $H$ divides the order of $G$, then $H$ is a subgroup of $G$." I'm sure you are able to find counterexamples to the latter statement easily.

The morale of the story is that "the converse" of an arbitrary statement is not well defined, and if it is, it could be nonsensical; even if the converse makes, the converse of a different but equivalent formulation of the same fact could be inequivalent to the original converse, and one can imagine one of them being true and another false. In the current case we have not found any "converse" that is actually true, but as an exercise you may try to find one... (A somewhat cheating trivial solution to this exercise is that, being a proven statement, Lagrange's Theorem is formally equivalent to every other proven statement, in the sense that assuming one of them you can prove the other. Now choose any proven statement whose converse happens to be proven as well.)

Lagranges theorem

$$H<G---->|G|/|H|$$

the converse says

$$|G|/|H|----->H<G$$

And the converse is not true, because A4 is a counter example and A4 is a group of oder 12 having no subgroup of order 6.

Proof:

A4 contains eight 3 cycles of the form (abc) and three pairs of disjoint transposition of the form (ab)(cd) and an identity element where a, b, c, d are disjoint elements.

If a sub group contains 3 cycles (abc) it must also contain it's inverse (acb).

If a sub group of order 6 exists then it must contain an identity element, a pair of transposition because the odd number of non identity elements cannot just be 3cycles (abc) and their inverses.

If A4 contains a sub group of order 6 it must contain at least two pairs of transposition because A4 contains only four elements that are not 3cycles.

And so without loss of generality:

Suppose the sub group of order 6 contains (abc) it must also contain(acb) it's inverse and since it contains a pair of transposition (ab)(cd) such that when we multiply the group is closed so (abc) (ab)(cd)=(bdc).

The inverse of (bdc)=(bcd)

(ab)(cd)(abc)=(acd)

(acd)invesre =(adc)

And then the identity elements which all together makes up more than 6 elements and so A4 has no subgroup of order 6.