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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?

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5  
Yes, it is true. –  Boris Novikov Feb 26 '13 at 14:10
2  
Yes, your "assumption" (in fact, deduction) is true. –  DonAntonio Feb 26 '13 at 14:10

4 Answers 4

up vote 7 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.

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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: $$\emptyset\not=\{e\}\subseteq H\cap I$$

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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.

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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.

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