# Union of more than $2$ subgroups can be a subgroup for some group

Actual Question is :

Suppose $H$ and $K$ are subgroups of a group $G$ such that $H\cup K$ is a subgroup of $G$. Prove that either $H\subseteq K$ or $K\subseteq H$

I could do this just by assuming that I have $h\in H; h\notin K$ and $k\in K; k\notin H$ then I would definitely have $hk\in H\cup K$ so $hk$ would be in either $H$ aor $K$ in any case it makes a contradiction.

Now the very next Question is :

Show that for each integer $n\geq 3$ there exists a group $G$ with subgroups $H_1,H_2,\cdots,H_n$ such that no $H_i$ is contained in any other and such that $H_1\cup H_2\cup \cdots \cup H_n$ is a subgroup of $G$.

I guess we have to construct some group for given $n\geq 3$

For some reason for $n=3$ I have the following idea :

$n=3$ so there would be three subgroups and no subgroup is contained in any other..

Each subgroup must have atleast $1$ non identity element so there are $3$ non identity elements adding up identity element i would get $4$ elements.

It would be extra ordinary if my choice of group is just of $4$ elements.

It would not be a wise choice if i choose my group to be cyclic so only choice i have is $\mathbb{Z}_2\times \mathbb{Z_2}=\{(0,0),(1,0),(0,1),(1,1)\}$

Each element is of $\mathbb{Z}_2\times \mathbb{Z_2}$ is of order $2$ so I would just take $H_1=\{(0,0),(1,0)\};H_2=\{(0,0),(0,1)\};H_3=\{(0,0),(1,1)\}$

So, I have $H_1\cup H_2\cup H_3$ to be whole group (which is More than sufficient)... I just need it to be a subgroup and it has become the whole group..

Now I got a group in which union of $3$ subgroups is a subgroup..

I have one more example for the case of $n=3$ - Quaternion Group

$H_1=\{\pm 1,\pm i\}; H_2=\{\pm 1,\pm j\};H_3=\{\pm 1,\pm ij\}$

Here also I have $H_1\cup H_2\cup H_3=\{\pm 1,\pm i,\pm j,\pm ij\}$ Which is a subgroup of $G$ and more over it is the whole group.

Now I have got another Question :

Is this just by chance that union of three proper subgroups is whole group of there are some examples in which union of three proper subgroups is a proper subgroup... I could not see this immediately

I really have no idea how to go with $n=4$ and so on...

Thank you.

• Your $k$ and $n$ in the second question are the same? – Singhal Feb 26 '14 at 7:49
• Yes.. I would edit that.. Thank you :) – user87543 Feb 26 '14 at 7:50
• If you take your examples of subgroups with union a group $G$, and choose another group $K$, all will be proper subgroups of $G \times K.$ – coffeemath Feb 26 '14 at 8:06
• @coffeemath : Yes Yes.. That does makes sense to me... I do not know what to say – user87543 Feb 26 '14 at 8:42
• So to do it for any $n$, use coffeemath's example $G \times K$ for some group $K$ with lots of subgroups. Then $G \times K$ is a union of three proper subgroups, and for the other subgroups, you can just choose $G \times L$ for lots of different subgroups $L$ of $K$. – Derek Holt Feb 26 '14 at 9:10

Derek Holt and Wei Zhou answered the question:

Show that for each integer $n\geq 3$ there exists a group $G$ with subgroups $H_1,H_2,\cdots,H_n$ such that no $H_i$ is contained in any other and such that $H_1\cup H_2\cup \cdots \cup H_n = G$.

(The original question specifies the union be a subgroup, but there is no loss in taking $G$ itself to be that subgroup.)

The idea is to take some smaller group $K$ that is the union of $n-1$ subgroups $K_1$, $K_2$, $\dots$, $K_{n-1}$, and then form $G=K \times L$ for some non-identity group $L$. Take $H_1=K_1 \times L$, $H_2=K_2 \times L$, $\dots$, $H_{n-1} = K_{n-1} \times L$ and $H_n = K \times 1$. You can verify that $H_n \not\subset H_i$ (since $H_i$ does not contain every $(x,1)$ for $x \in K$) and $H_i \not\subset H_n$ (since $H_n$ does not contain $(1,x)$ for any $1\neq x \in L$).

However, Praphulla Koushik was understandably confused since $G$ was also the union $H_1 \cup \dots \cup H_{n-1}$, the last $H_n$ was redundant. I don't address the existence of non-redundant unions (or “coverings”) of size $n$, but I do address the minimal size of a covering:

Let $\sigma(G)$ be the least $n$ such that $G$ is the union of $n$ proper subgroups.

Clearly $\sigma(C_n) = \infty$ but otherwise for a finite group $G$, $\sigma(G)$ is finite. The obvious generalization is:

Show that for each integer $n$ there exists a group $G$ with $\sigma(G)=n$.

However, Cohn (1994) conjectured this was impossible for $n=7$, and Tomkinson (1997) proved this. Tomkinson also gave a formula for $\sigma(G)$ when $G$ is solvable: $\sigma(G)=[H:K]$ where $H/K$ is a the smallest chief factor with more than one complement.

We say that a group $G$ is $\sigma$-primitive if $G$ has no non-identity normal subgroup $N$ with $\sigma(G)=\sigma(G/N)$. In some sense, the $\sigma$-primitive groups are the only interesting ones (as the others just have $N \leq \cap H_i$ so the $N$ part could have been ignored).

As a standard exercise, $\sigma(G)\leq 2$ is impossible.

Cohn found the $\sigma$-primitive groups $G$ with $\sigma(G)=3$: only $C_2 \times C_2$, the same group Praphulla Koushik deduced. He also handled $\sigma(G)=4$: only $C_3\times C_3$ and $S_3$, and $\sigma(G)=5$: only $A_4$. Notice each of these is of the form $p^a+1$ for $p$ prime and $a$ positive. The first number $\geq 3$ not of this form is $7$, and it is not a $\sigma$-number. Current research involves studying the structure $\sigma$-primitive groups and finding $\sigma(G)$ for groups $G$ known to be $\sigma$-primitive. Detomi–Lucchini (2008) gives a good indication on the structure, and several papers compute $\sigma$ for (usually finitely many) simple groups.

## Bibliography

• So I am still wondering whether for every $n\ge 3$ we can find a group that is a non-redundant union of $n$ subgroups. That's easy for $n=7$: the subgroups of order $2$ in an elementary group of order $8$. – Derek Holt Feb 26 '14 at 17:30
• Yes, I think non-redundant is an interesting question; I just didn't have any good answer. I've attended a few talks on $\sigma$, and so found it easy to write up this little bit (haven't had time to find links to online copies of the papers yet). – Jack Schmidt Feb 26 '14 at 18:20
• – Jack Schmidt Feb 26 '14 at 18:20
• I was sure you would understand my problem and you did... I could not understand why would you want to pick another subgroup (adding upto $n$ subgroups) when the group is already being covered by $n-1$ subgroups... Now it does makes sense to me... This problem is much harder than i thought it would be :D – user87543 Feb 27 '14 at 5:05
• @PraphullaKoushik: On the contrary, the question you actually asked was easy, which seemed to be what was confusing you! There are some much more difficult related questions. – Derek Holt Feb 27 '14 at 8:53

Suppose you have a "$k$"-example: $G=A_1 \cup \cdots \cup A_k$, where $A_i <G$. Let $H$ be a group isomorphic to $G$, then there exists $B_1 \le H$ and $B_1 \cong A_1$. Now consider the group $X=G \times H$. Clearly $X=A_1 \times H\cup \cdots \cup A_k \times H$. Let $X_i=A_i \times H$. So $X=X_1 \cup \cdots X_k$. Let $X_{k+1}=G \times B_1$. Now you can find that $X, X_1, \cdots, X_{k+1}$ is a "$k+1$"-example. So if $k$ is right, then $k+1$ is right.

• I do not really get your idea behind choosing $A_{k+1}=G\times A_1$ :O – user87543 Feb 26 '14 at 9:17
• Maybe it is easy to understand if we write the two isomorphic direct factors in two different letters. – Wei Zhou Feb 26 '14 at 10:49
• I have reedited it. – Wei Zhou Feb 26 '14 at 11:04
• Why not use $H=G$ and $B_1=A_1$? – Christoph Feb 26 '14 at 11:15
• I am getting confused with the notation... Let me tell you what i have understood.... Suppose for $k\geq 3$ I have a group $G$ such that $G=H_1\cup H_2\cup \cdots \cup H_k$.. Now i have to find another group such that there are $k+1$ subgroups whose union is a subgroup... Another reason i am getting confused is I was expecting union to be a subgroup (possibly proper) but it is actually whole group.. – user87543 Feb 26 '14 at 11:45