This is a true or false question, hence are the answers supposed to follow quickly? Because the empty set has no identity element, hence $\emptyset$ is not a group. Hence I'm inquiring for intersection and union $\neq \emptyset$. Because the intersection of subgroups is a subgroup, I guessed intersection of groups is truly a group? But Fraleigh's answer says false?

I inquired about the union of subgroups is not a subgroup, but what about for groups? I don't know how to predestine, preordain this because that other question, for sub groups, still feels "fatidic" or magical to me. To boot, the difference, for the intersection of subgroups vs that of groups, confounds me as to what to do. What's the intuition?

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    $\begingroup$ Two groups needn't even share the same binary operation. What binary operation do you propose to keep on the intersection? And what about the identity element? Clearly, intersection/union of groups needn't be a group. With respect to subgroups, the intersection is always a subgroup. The union needn't be a subgroup. $\endgroup$ – Suhas M Apr 26 '14 at 13:48

The intersection of two groups $(U,\odot_U)$ and $(V,\odot_V)$ is first, and foremost, the set $$ G = U \cap V \text{.} $$ To turn this into a group, one would need to define a suitable operation $\odot_G$ on $G$. But where is that operation supposed to come from? Since $U$ and $V$ can be completely different groups, which just happen to be constructed over two non-disjoint sets $U$ and $V$, it's not at all obvious how $\odot_G$ is supposed to be defined. Thus, the intersection of two groups is merely a set, not a group. The same holds of course for the union of two sets - again, where would the operation come from that turns the union into a group?

In fact, since we generally consider groups only up to isomorphisms, i.e we treat two groups $G_1,G_2$ as the same group $G$ if they only difference between the two is the names of the elements, the union or intersection of two groups isn't even well-defined. For any pair of groups $U,V$ we can find some set-theoretic representation of $U$ and $V$ such that $U \cap V = \emptyset$, and another such that $U \cap V \neq \emptyset$.

Now constrast this with the situation of two subgroups $U,V$ of some group $(H,\odot_H)$. In this case, we know that $\odot_U$ and $\odot_V$ are simply the restrictions of $\odot_H$ to $U$ respectively $V$, and the two operations will therefore agree on the intersection of $U$ and $V$. So we can very naturally endow the set $$ G = U \cap V $$ with the operation $$ \odot_G = \odot_H\big|_{U \cap V} = \odot_U\big|_{U \cap V} = \odot_V\big|_{U \cap V} \text{.} $$

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    $\begingroup$ The first disaster with intersection might in fact be that $U\cap V=\emptyset$ $\endgroup$ – Hagen von Eitzen Apr 26 '14 at 13:49
  • $\begingroup$ @HagenvonEitzen Yeah, I realized after posting that I should mention that the intersection is not particularly well-defined. Edited to say that. $\endgroup$ – fgp Apr 26 '14 at 13:56

There is one important special case (well, it's the only valid case, so not really special). Assume you have a family of groups, $\{G_i\}_{i\in I}$, indexed by some nonempty totally ordered index set $I$. Also assume that $G_i$ is a subgroup of $G_j$ whenever $i\le j$. Then on the set-theoretic union $G:=\bigcup_{i\in I} G_i$ we obtain a unique group structure such that each $G_i$ is a subgroup of $G$. To see this note that we can - and must - define $g *_G h$ as follows for $g,h\in G$: There exist $i,j\in I$ with $g\in G_i$, $h\in G_j$. Then $g,h\in G_k$ for all $k\ge \max\{i,j\}$ and we can and must define $g*_Gh=g*_{G_k}h$ for any $k\ge \max\{i,j\}$.


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