The Intersection and Union of Two Subgroups

# The Intersection and Union of Two Subgroups

Consider a group $(G, \cdot)$ and suppose that $(S, \cdot)$ and $(T, \cdot)$ are both subgroups of $(G, \cdot)$. What can we then say about the set intersection $S \cap T$ and set union $S \cup T$ with respect to the operation $\cdot$? Will they always form groups or not? The following propositions answer these questions.

Proposition 1: Let $(G, \cdot)$ be a group and $(S, \cdot)$ and $(T, \cdot)$ be subgroups of $(G, \cdot)$. Then $(S \cap T, \cdot)$ is also a subgroup of $(G, \cdot)$. |

**Proof:**Since $(S, \cdot)$ and $(T, \cdot)$ are subgroups of $(G, \cdot)$ we have that $S, T \subseteq G$ and so $S \cap T \subseteq G$. We therefore only need to show that $S \cap T$ is closed under $\cdot$ and that for each $a \in S \cap T$ there exists an $a^{-1} \in S \cap T$ such that $a \cdot a^{-1} = e$ and $a^{-1} \cdot a = e$ where $e \in G$ is the identity with respect to $\cdot$.

- Let $x, y \in S \cap T$. Then $x, y \in S$ and $x, y \in T$ and since $(S, \cdot)$ and $(T, \cdot)$ are groups themselves, we have that $S$ and $T$ are both closed under $\cdot$. Therefore $(x \cdot y) \in S$ and $(x \cdot y) \in T$. Therefore $(x \cdot y) \in S \cap T$, so $S \cap T$ is closed under $\cdot$.

- Now let $a \in S \cap T$. Then $a \in S$ and $a \in T$. Since $(S, \cdot)$ and $(T, \cdot)$ are groups, we have that there exists $a^{-1} \in S$ such that $a \cdot a^{-1} = e$ and $a^{-1} \cdot a = e$ and there exists $a^{-1'} \in T$ such that $a \cdot a^{-1'} = e$ and $a^{-1} \cdot a = e$. Since $S, T \subseteq G$ we must have that $a^{-1} = a^{-1'}$ otherwise this would contradict the 'existence of a unique inverse for $a \in G$. Therefore $a^{-1} \in S$ and $a^{-1} \in T$ so there exists an $a^{-1} \in S \cap T$ such that $a \cdot a^{-1} = e$ and $a^{-1} \cdot a = e$.

- Therefore $(S \cap T, \cdot)$ is a group. $\blacksquare$

Proposition 2: Let $(G, \cdot)$ be a group and let $(S, \cdot)$ and $(T, \cdot)$ be subgroups of $(G, \cdot)$. Then $(S \cup T, \cdot)$ is a group if and only if $S \subseteq T$ or $S \supseteq T$. |

**Proof:**$\Rightarrow$ Suppose that $(S \cup T, \cdot)$ is a group. Then all of the group axioms hold on $S \cup T$ with respect to the operation $\cdot$ and so for all $x, y, z \in S \cup T$ we have that $(x \cdot y) \in S \cup T$; $x \cdot (y \cdot z) = (x \cdot y) \cdot z$; there exists an $e \in S \cup T$ such that $x \cdot e = x$ and $e \cdot x = x$; and $x \in S \cup T$ implies that there exists $x^{-1} \in S \cup T$ such that $x \cdot x^{-1} = e$ and $x^{-1} \cdot x = e$.

- Now suppose that $S \not \subseteq T$ and $S \not \supseteq T$. Then there exists an $x \in S$ such that $x \not \in T$ and similarly there exists a $y \in T$ such that $y \not in S$. Clearly $x, y \in S \cup T$ so $(x \cdot y) \in S \cup T$ since $(S \cup T, \cdot)$ is a group (and hence closed under $\cdot$). So either $(x \cdot y) \in S$ or $(x \cdot y) \in T$.

- Suppose that $(x \cdot y) \in S$. Since $x \in S$ and $(S, \cdot)$ is a group, we have that $x^{-1} \in S$. Since $(S, \cdot)$ is a group, we also have that $S$ is closed under $\cdot$ and so:

\begin{align} \quad x^{-1} \cdot (x \cdot y) = (x^{-1} \cdot x) \cdot y = e \cdot y = y \in S \end{align}

- But this is a contradiction since $y \not \in S$.

- Similarly, suppose that $(x \cdot y) \in T$. Since $y \in T$ and $(T, \cdot)$ is a group, we have that $y^{-1} \in T$. Since $(T, \cdot)$ is a group, we also have that $T$ is closed under $\cdot$ and so:

\begin{align} \quad (x \cdot y) \cdot y^{-1} = x \cdot (y \cdot y^{-1}) = x \cdot e = x \in T \end{align}

- But this too is a contradiction since $x \not \in T$.

- Therefore the assumption that $S \not \subseteq T$ and $S \not \supseteq T$ was false. Therefore either $S \subseteq T$ or $S \supseteq T$.

- $\Leftarrow$ Suppose that now $S \subseteq T$. Then $S \cup T = T$ and since $(T, \cdot)$ is a group, we have that $(S \cup T, \cdot)$ is a group. Now suppose that $S \supseteq T$. Then $S \cup T = S$ and since $(S, \cdot)$ is a group, we once again have that $(S \cup T, \cdot)$ is a group. $\blacksquare$