How to prove that the converse of Lagrange's theorem is not true?

Solution 1:

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.

Solution 2:

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 $20=2^2\cdot 5$, 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 $p\ \text{mod}\,q$ and $f$ be the exponent of $q\ \text{mod}\,p$. 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 $45=3^2\cdot 5$ or $135=3^3\cdot 5$ 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.)