Return to Theorems, Glossary, Homework Problems.

**Statement:**

Let $G$ be a group, $H$ a subgroup of $G$, and $a,b \in G$. Then:

- $a \in aH$
- $aH = H \iff a \in H$
- Either $aH = bH$ or $aH \cap bH = \emptyset$
- $aH = bH \iff a^{-1} b \in H$
- $aH \leq G \iff a \in H$
- $|aH| = |bH| = |H|$

**Proof:**

1. Since $H$ is a subgroup of $G$, $e \in H$, so $a = ae \in aH$.

2. We must prove this statement both ways since it is an if and only if statement. ($\Rightarrow$) Suppose $aH = H$; we need to show that $a \in H$. From Property #1, we know $a \in aH$, but $aH = H$, so $a \in H$. ($\Leftarrow$) Suppose $a \in H$; we need to show that $aH = H$. In order to do this, we need to show $aH \subseteq H$ and $H \subseteq aH$. First, let $x \in aH$ and define $x = ah$ for some $h \in H$. Since $H$ is a subgroup, we know that it is closed under its operation, and since $a,h \in H, x = ah \in H$ Therefore $aH \subseteq H$. Now let $x \in H$. Since $H$ is a subgroup, we know that it contains the $e$ and $a^{-1} \forall a \in H$. So, $x = ex = aa^{-1}x = a(a^{-1}x)$. Since $a^{-1}$ and $x$ are both in $H$, $a^{-1}x \in H$ as well. Therefore, $x = a(a^{-1}x) \in aH$ making $H \subseteq aH$. With $aH$ and $H$ being subsets of each other, it follows that $aH = H$.

3. We will have either $aH \cap bH = \emptyset$ or they will share at least one element. Let that one elements be called $c$, i.e. $c \in aH \cap bH$. If we can show that if $c \in aH \implies cH =aH$, then we can show that $aH = bH$. To do this, we must show that $cH$ and $aH$ are subsets of each other. Suppose $c \in aH$ and let $x \in cH$. Then, $x = ch_1$ for some $h_1 \in H$. However, since $c \in aH$, $c = ah_2$ for some $h_2$ in $H$. So, $x = ah_{2}h_{1}$ Since $h_{2}h_{1} \in H$ (because $H$ is closed), we can see that $x \in aH$, so $cH \subseteq aH$. Now let $x \in aH$. Then, $x = ah_1$ for some $h_1 \in H$. But since $c \in aH$, we know $c = ah_2$ for some $h_2 \in H$ which implies $a = ch_{2}^{-1}$ so then $x = ch_{2}^{-1}h_{1}$ which is an element of $cH$, making $aH \subseteq cH$. Since both $aH$ and $cH$ are subsets of each other, $aH = cH$. Now that we know $c \in aH \implies cH = aH$, we can use it for both $aH$ and $bH$. So, since $c \in aH \cap bH$, $cH = aH$ and $cH = bH$ which implies $aH = bH$.

4. From Property #3, we proved that if $a \in bH$ then $aH=bH$. Consider the converse of this statement: if $aH = bH$, then $a \in bH$. This turns out also to be true. Suppose $aH = bH$. Since $H$ is a subgroup, it contains $e$, so $a = ae \in aH$. But, since $aH = bH$, $a = ae \in aH = bH$, hence $a \in bH$. Now that we have proved the converse as well as the original statements, it can be written as an if and only if statement, i.e. $aH = bH \iff a \in bH$. Therefore, $bH = aH \iff b \in aH$ which can be written in the form $b = ah$ for some $h \in H$. Also, we know that $H$ contains both the identity and inverse, so $b = ah \iff a^{-1}b = a^{-1}ah \iff a^{-1}b = h \iff a^{-1}b \in H$.

5. We need to prove this statement both ways since it is an if and only if statement. ($\Rightarrow$) Suppose $aH \leq H$; we need to show that $a \in H$. Since $aH$ is a subgroup of $G$, we know that $e \in aH$ which can be written in the form $e = ah$ for some $h \in H$. We also know that $aH$ contains the inverse, so we can multiply both sides by $h^{-1}$ on the right and get $eh^{-1} = ahh^{-1} \implies h^{-1} = ae \implies h^{-1} = a$. Hence $a = h^{-1} \in H$. ($\Leftarrow$) Suppose $a \in H$; we need to show that $aH \leq G$. From Property #2, we know that if $a \in H$, then $aH = H$. So, $a \in aH = H \implies a \in H$ and $H$ is a subgroup of $G$.

6. In order to show the order of two sets to be equal, we need to establish a one-to-one function from one set onto another. Let $\phi: aH \rightarrow bH$ be defined as $\phi(ah) = bh$ for some $h \in H$. Elements in $aH$ are in the form $ah$ for some $h \in H$. Similarly, elements in $bH$ are in the form $bh$ for some $h \in H$. Let $x = ah$ and $y = bh$. Let's check to see if $\phi$ is one-to-one. Suppose $\phi(x_1) = \phi(x_2)$. Then we have $bh_1 = y_1 = y_2 = bh_2$, which shows $\phi$ is one-to-one. Now we need to check to see if $\phi$ is onto. This means that $\forall y \in bH \; \exists x \in aH$ such that $\phi(x) = y$. This is easy to see, just let $x = x \in aH$. We then have $\phi(x) = \phi(ah) = bh = y$. We have shown that $\phi$ is onto as well. Since we have established a one-to-one correspondence between $aH$ and $bH$, we can say that they have the same order, i.e. $|aH| = |bH|$.