Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is there any group $(G,+)$, that has $2$ or more subgroups $(H,+),\; (I,+)$, where

$$H \cap I = \emptyset?$$

I believe not, because both $H$ and $I$ must contain neutral element.

Is my assumption true?

share|cite|improve this question
Yes, it is true. – Boris Novikov Feb 26 '13 at 14:10
Yes, your "assumption" (in fact, deduction) is true. – DonAntonio Feb 26 '13 at 14:10
up vote 8 down vote accepted

You are completely correct. Any subgroup must contain the neutral (identity) element, so their intersection must also contain this element, and is thus not empty.

Note, however, that one will often say that two subgroups intersect trivially if the intersection contains only the neutral element.

share|cite|improve this answer

You are right, two subgroups have always the neutral element $e$ in common. This might well be the only element in common even if the two subgroup are both different from $\{ e \}$, see e.g. $S_{3}$, or the Klein $4$-group.

share|cite|improve this answer

Let's try to prove this out of the axioms of a group. We know that $G,H,I$ are all groups under $+$ and because $H,I$ are subgroups we know that $H,I\subseteq G$.

Let $h\in H$ and $i\in I$, then $h,i\in G$. By defintion there exist an element $e\in G$, such that
\begin{align} &e+h=h \tag 1\\ &e+i=i \tag 2 \end{align}

But as $H,I$ are also groups, there exist also an element $e_h\in H,e_i\in I$ such that
\begin{align} &e_h+h=h \tag 3\\ &e_i+i=i \tag 4 \end{align}

By definition of a group, $h$ and $i$ both have a left-inverse, therefore we obtain: \begin{align} &e+h\overset{1}{=}h\overset{3}{=}e_h+h\implies e=e_h \\\\ &e+i\overset{2}{=}i\overset{4}{=}e_i+i\implies e=e_i \end{align}

This implies your result: $$\emptyset\not=\{e\}\subseteq H\cap I$$

share|cite|improve this answer

Yes, you're correct, every subgroup must contain the identity element of $G$, hence the intersection of any two subgroups of a group must necessarily be non-empty: at the very least, if $H_1\leq G, H_2 \leq G,$ then $H_1 \cap H_2 \leq G$, even if the only element in common is the neutral/identity element.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.