Let $H$ and $K$ be two subgroups of a group $G$. Then $H \cup K$ is a subgroup $\iff H\subset K$ or $K \subset H$.

My proof: (probably contain errors)

If one subgroup is a subset of other then $H\cup K$ is a subgroup. The proof of this implication is easy.

Now if $H\cup K$ is a subgroup, we have to prove $H\subset K$ or $K \subset H$.

Let $x\in H$ and $y\in K$

Then $x*y \in H \cup K$ for $H \cup K$ is a subgroup.

$\implies x*y \in H $ or $x*y \in K$

$\implies y\in H $ or $ x \in K$

If $y\in H$, then $K\subset H$. If $y \notin H$, then $x \in K$ which implies $H \subset K$

$\therefore$ $H\subset K$ or $K \subset H$.

I know that this proof is wrong. Please help me improve the proof. Suggestions and hints are welcomed. Preference is given to direct proof which uses only the elementary properties of a group (proof which help me to improve the proof I had written) but proof by contradictions are also okay. Thanks in advance.

  • 2
    $\begingroup$ $x*y\in H\cap K$ is false. What is true is that it's in $H\cup K$. $\endgroup$
    – Bernard
    Aug 8, 2017 at 14:05
  • $\begingroup$ How do you get from $y\in H$ to $K\subset H$? $\endgroup$ Aug 8, 2017 at 14:07
  • $\begingroup$ @Lord sorry I know it is wrong. I started with a $y \in K$ and arrived at a conclusion that $y \in H$. My idea is that suppose $x_1,x_2 \in H$ and $y_1, y_2 \in K$ such that $x_1 \in K $ but $x_2 \notin K$. But this forces both $ y_1$ and $y_2 \in H$. $\endgroup$ Aug 8, 2017 at 14:18
  • $\begingroup$ @Lord I have this rotten proof in my hand. I want to improve this $\endgroup$ Aug 8, 2017 at 14:21
  • $\begingroup$ The most glaring error in my opinion is saying "...this implication is easy". If you're asking for people to check your work, then actually spell out the details, easy or not, or else don't bother writing it at all. $\endgroup$
    – Junglemath
    Mar 23, 2020 at 16:15

2 Answers 2


We will prove that

$(1)\quad K\cup H $ is a subgroup $\implies K\subset H \lor H\subset K$

By contradiction

$ (H\not\subset K)\land (K\not\subset H) \iff \left\lbrace\begin{array}l \exists h\in H\backslash K \\\exists k\in K\backslash H \end{array}\right.$

Let $u= k*h\in K\cup H$

$k*h\in K\cup H \implies \left\lbrace \begin{array}l k*h\in K\implies k^{-1}*k*h\in K\implies h\in K \quad\text{contradiction}\\\text{or}\\k*h\in H\implies k*h*h^{-1}\in H\implies k\in H \quad\text{contradiction}\end{array}\right.$

So $K\cup H $ is a subgroup $\implies K\subset H \lor H\subset K$

$ (2)\quad K\subset H \lor H\subset K \implies $ $K\cup H $ is a subgroup is obvious

  • $\begingroup$ Any direct proof please $\endgroup$ Aug 8, 2017 at 14:37
  • 2
    $\begingroup$ The first one should be $k^{-1}*k*h = h$. $\endgroup$
    – platty
    Aug 8, 2017 at 14:37
  • $\begingroup$ proof which help me to improve the proof I had written $\endgroup$ Aug 8, 2017 at 14:38
  • $\begingroup$ @platty thanks and done $\endgroup$
    – Stu
    Aug 8, 2017 at 14:45

I think any such proof would revolve around the same ideas. But here's how we can phrase it to stay close enough to your original proof.


… we have to prove $H\subset K$ or $K\subset H$

we can insert the following: If $H\subset K$, we're done. So let's assume that $H\not\subset K$ (in which case our goal is to show that $K\subset H$). Then instead of

Let $x\in H$

it's better to pick $x\in H\setminus K$. Now we can pick any $y\in K$, as you did. And as in your proof (but I just want to give a name to the new element):

Then $z=x*y\in H\cup K$ for $H\cup K$ is a subgroup $\implies$ $x*y\in H$ or $x*y\in K$.

If $z=x*y\in K$, then $x=zy^{-1}\in K$, since both $z\in K$ and $y^{-1}\in K$. But this is impossible because we chose $x\in H\setminus K$. Therefore $z=x*y\in H$. But then $y=x^{-1}z\in H$, since both $x^{-1}\in H$ and $z\in H$.

Thus we've shown that for any element $y\in K$, we have $y\in H$, i.e. $K\subset H$.


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