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Given $G$ is a group with $H$ & $K$ subgroups of $G$, find an example such that $H\cup K$ is not a subgroup of $G$.

I'm having trouble trying to come up with a suitable example, I can outline the proof of why $H\cup K$ is not necessarily a subgroup, i.e I understand that given $H\nsubseteq K$ and $K\nsubseteq H$ then $H\cup K$ can not satisfy closure.

However I'm having trouble imagining two subgroups such that one is not contained inside the other and both still contain the identity.

I'm not looking for someone to give me an example, just to be pointed in the right direction.

Thanks

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You can find examples in any finite group that is not cyclic. –  spin Mar 2 at 14:40

3 Answers 3

up vote 2 down vote accepted

There are many examples on this: Consider the group $(\mathbb{Z}, +)$. Now $(3\mathbb{Z}, +)$ and $(4\mathbb{Z}, +)$ are its subgroups, but is their union also a subgroup of $(\mathbb{Z}, +)$?

The other example is, consider the set $X$ of all vectors on the plane $Oxy$, with the normal addition operation; and given 2 arbitrary lines $d_1$, and $d_2$ that cut each other, then the sets of all vectors whose supports are parallel to $d_1$, and $d_2$ respectively are subgroups of $X$, whereas their union is clearly not.


And in fact, you can prove the following proposition.

Proposition

Given the group $X$, and $H$, $K$ are its subgroups, then for $H \cup K$ to be a subgroup of $X$, it's necessary, and sufficient that either $H$ contains $K$, or $K$ contains $H$. (Try to prove this asertion, it should be pretty easy, one way is clear, and the other way should be solved by a simple Proof by Contradiction).

So, from the above proposition, all subgroups $H$, and $K$ of $X$ in which $H$ does not contain $K$, and $K$ does not contain $H$ either, will satisfy the requirement that $H \cup K$ is not a subgroup of $X$.

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Look at non-cyclic groups, e.g. look at the Klein-4 group, or $S_3$, and its subgroups.

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Consider the group $\mathbb{Z}^2$. Let $H=\{(x,0):x\in\mathbb{Z}\}$ and $K=\{(0,x):x\in\mathbb{Z}\}$. Note that $H$ and $K$ are subgroups of $\mathbb{Z}^2$. However, if we take $(1,0)\in H$ and $(0,1)\in K$, we find their sum $(1,1)$ is not in $H\cup K$. Therefore, $H\cup K$ is not a subgroup.

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