Date modified: Thu 2023-05-11 . 11:51 AM
Date created: Sat 2023-10-07 . 09:02 PM
# When is a group a union of 3 proper subgroups? Let $G$ be a group. Then $G$ is a union of three proper subgroups if and only if there exists some normal subgroup $H$ such that $G/H\approx V_4$, the Klein-4 group. --- Remark / Warm-up. Note, **no group can be written as a union of two proper subgroups**. To see this, if to the contrary that $G=H_1 \cup H_2$ for some two proper subgroups $H_1, H_2$ of $G$, then there exists some elements $a\in H_1 - H_2$ and $b \in H_2 - H_1$. Consider the element $ab$, which is neither in $H_1$ nor $H_2$, as subgroups are closed under products and inverses. --- Alright, let's do one direction at a time. Easy direction first, maybe. Suppose there exists some normal subgroup $H\trianglelefteq G$ such that $G/H\approx V_4 = \{1,a,b,c\}$. Then this means there is a surjective homomorphism $\phi : G \to V_4$ with $\ker \phi = H$. Now $V_4$ has three natural proper sugroups: $A=\{1,a\}$, $B=\{1,b\}$, and $C=\{1,c\}$, where $V=A\cup B \cup C$. Consider then the preimages $\phi^{-1}(A)$, $\phi^{-1}(B)$, and $\phi^{-1}(C)$. Each of these is a subgroup of $G$, and they must be proper, as $\phi$ is surjective. And by elementary fact of sets, that $G = \phi^{-1}(A\cup B\cup C) = \phi^{-1}(A) \cup \phi^{-1}(B) \cup \phi^{-1}(C)$, union of three proper subgroups! --- Now, suppose $G = H_1 \cup H_2 \cup H_3$, a union of three proper subgroups. Since no two proper subgroups can union to $G$, each $H_i$ contains something that is not present in the union of the other two. Now diagramatically we have the following, with what we know so far: ![[---images/--- When is a group a union of 3 proper subgroups 2023-05-03 22.32.39.excalidraw.svg]] Let us now look at the intersection $H_1 \cap H_2$. Take $x\in H_1 \cap H_2$. Suppose $x$ is not in $H_3$. Note there is an element $z$ in $H_3$ not from $H_1$ nor $H_2$. What can we say about $xz$? If $xz \in H_1$, then $z\in H_1$, contradiction. If $xz \in H_2$, then similarly $z\in H_2$, a contradiction. Lastly, if $xz\in H_3$, then $x\in H_3$, also a contradiction! Hence $x$ is an element in $H_3$. In other words, $H_1\cap H_2 \subset H_3$. And similarly $H_1\cap H_3\subset H_2$, and $H_2\cap H_3 \subset H_1$. So! Our picture of $G$ loos more like this: ![[---images/--- When is a group a union of 3 proper subgroups 2023-05-03 22.45.15.excalidraw.svg]] Where each of the four pieces is not empty (in particular $e\in H_1\cap H_2\cap H_3$). Now, to show $G$ has a quotient that is the Klein-4 group $V_4$ is to show a homomorphism surjective onto $V_4=\{1,a,b,c\}$. The most natural way is to define such a map $\phi : G\to V_4$ as follows: ![[---images/--- When is a group a union of 3 proper subgroups 2023-05-03 22.52.13.excalidraw.svg]] So it remains to show this is indeed a group homomorphism. Take two elements from different color regions. Take $x \in H_1 - (H_2\cup H_3)$ and $y \in H_2 - (H_1\cup H_3)$. Note $\phi(x) = a$ and $\phi(y)=b$. Now, if $xy$ is in $H_1$, then $y$ is in $H_1$, a contradiction. If $xy$ is in $H_2$, then $x$ is in $H_1$, contradiction. So $xy$ is in $H_3 - (H_1\cup H_2)$. Whence $\phi(xy) = c$, which agrees with $\phi(x)\phi(y)=ab=c$! This gives the three symmetric cases: Two elements from different colors have a product in the third color. Let us quickly check the other essentially different cases. Two elements from the same color: Take $x,x' \in H_1 -(H_2 \cup H_3)$. Certainly $xx' \in H_1$. Suppose that $xx'\in H_1-(H_2\cup H_3)$. Take $y \in H_2 - (H_1\cup H_2)$. Then from previous we know $yx\in H_3-(H_1\cup H_2)$. And hence $(yx)x'\in H_2 - (H_1\cup H_3)$. But $(yx)x' = y(xx') \in H_3 - (H_1\cup H_2)$, a contradiction! Hence $xx'$ is in one of $H_2$ or $H_3$. But by picture we know this implies $xx' \in H_1\cap H_2\cap H_3$. Hence $\phi(xx')=1=aa=\phi(x)\phi(x')$. So, two elements from the same color region have a product in the white region. Lastly, one element from a color region, and the other in the white region. Say $x\in H_1 - (H_2\cup H_3)$ and $w\in H_1\cap H_2 \cap H_3$. Note $xw$ is in $H_1$. Suppose $xw \in H_2$, then $x\in H_2$, a contradiction. Similarly if $xw\in H_3$, then $x\in H_3$, also contradiction. Hence $xw\in H_1-(H_2\cup H_3)$! Hence $\phi$ indeed defines a surjective homomorphism, and thus $G$ has $V_4$ as a quotient! Neat! --- For some reason, I find this result remarkable. By the way, in this case, $G$ has a **flower arrangement of subgroups**. Can we say more about groups like this? Curiously, we have the following results. Denote $\sigma(G)=n$ to mean $G$ can be written as a union of $n$ many proper subgroups, but no fewer. Then $\sigma(G)=3$ $\iff$ $G$ has a quotient of $V_4$. $\sigma(G)=4$ $\iff$ $\sigma(G)\neq 3$ and $G$ has a quotient of $S_3$ or $C_3 \times C_3$. $\sigma(G)=5$ $\iff$ $\sigma(G)\neq 3,4$ and $G$ has a quotient of $A_4$. $\sigma(G)=6$ $\iff$ $\sigma(G)\neq 3,4,5$ and $G$ has a quotient of $Dih_5$, $C_5 \times C_5$, or $W=C_4 \ltimes C_5$, where $W$ is the group of order 20 generated by $a,b$ with relations $a^5=b^4=e$, $ba=a^2b$. The $\sigma(G)=3$ result here is by Scorza, then Cohn worked out the $\sigma(G)=4,5,6$ cases. What about $7$? And curiously Tomkinson showed: There is no group $G$ with $\sigma(G)=7$. Interesting... #group-theory #klein-4-group