# $G$ is a finite group and $G_1$,$G_2$ are subgroup that $G_1 \cap G_2=\{1\}$, then $|G_1|.|G_2| \big| |G|$?

$G$ is a finite group and $G_1$,$G_2$ are subgroup that $G_1 \cap G_2=\{1\}$, then $|G_1|.|G_2| \big| |G|$?

If $x\in G_1$, $y \in G_2$ implies $xy=yx$ or one is normal, the statement is obvious. But other cases, I can't do anything. Is there some weaker condition to prove it, or there are some counterexamples?

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Let $r_1,r_2 \in D_n$ two different reflections in the dihedral group of order $2n$. Take $G_1= \langle r_1 \rangle$ and $G_2 = \langle r_2 \rangle$. Then $G_1 \cap G_2= \{1\}$ but $|G_1 | \cdot |G_2|=4$ doesn't divide $|D_n|=2n$ if $n$ is odd.

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For a weaker condition, this is true when $G$ is nilpotent. We can show the following

A finite group $G$ is nilpotent if and only if $|G_1G_2|$ divides $|G|$ for every subgroup $G_1$, $G_2$ of $G$.

If $G$ is nilpotent, then $G$ is a direct product of its Sylow subgroups. That is, $G = P_1 \times \ldots \times P_t$ where $P_i$ are the Sylow subgroups of $G$. Then $G_1$ and $G_2$ are nilpotent as well, so $G_1 = A_1 \times \ldots \times A_t$ and $G_2 = B_2 \times \ldots \times B_t$ where $A_i$ and $B_i$ are subgroups of $P_i$. Now the product $G_1G_2$ is not necessarily a subgroup, but it is of the form

$$G_1G_2 = A_1B_1 \times \ldots \times A_tB_t$$

so $|G_1G_2|$ divides $|G|$ because each $|A_iB_i|$ divides $|P_i|$.

For the other direction, note that for two Sylow $p$-subgroups $P$ and $Q$, the order $|PQ|$ divides $|G|$ if and only if $P = Q$. Hence for each prime divisor of $G$ there is exactly one Sylow subgroup, which implies that $G$ is nilpotent.

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+1 Truly nice one! –  Andreas Caranti Feb 9 '13 at 15:14

More generally (than the case Seirios gives), let $G$ be a finite group of order divisible by the prime $p$, and suppose that $G$ has no non-identity normal $p$-subgroup and also has an Abelian Sylow $p$-subgroup $P.$ Then by a theorem of Brodkey, there is some $g \in G$ such that $P \cap g^{-1}Pg = 1$, but we certainly don't have $|P|^{2}$ dividing $|G|$.

By the way, it may be worth noting that the question has a positive answer if $G$ itself is a $p$-group for some prime $p$, as then $|G_{1}G_{2}|$ is a power of $p$ ( though $G_{1}G_{2}$ need not be a subgroup) and this power of $p$ is at most $|G|$, which is also a power of $p.$

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