We see that cyclic group of order $d$ looks very similar to the group $\left(\mathbb{Z}_{d},+\right) .$ In some sense this is the same group, and to make this claim precise we introduce a new notion.
A group isomorphism between groups $\left(G_{1}, *{1}\right)$ and $\left(G{2}, *_{2}\right)$ is a bijection
$$
\varphi: G_{1} \rightarrow G_{2}
$$
which respects operations ${1}$ and ${2}$, i.e., for any $x, y \in G_{1}$ we have
$$
\varphi\left(x *{1} y\right)=\varphi(x) *{2} \varphi(y) .
$$
Let $\varphi: G_{1} \rightarrow G_{2}$ be an isomorphism. Then $\varphi^{-1}: G_{2} \rightarrow G_{1}$ is also an isomorphism.
Proof. Since $\varphi$ is a bijection, $\varphi^{-1}$ is also a bijection.
It remains to check that $\varphi^{-1}$ respects multiplication. Let $a, b, \in G_{2}$ be any two elements. Since $\varphi$ is a bijection, we can uniquely write
$$
a=\varphi(x) \quad b=\varphi(y)
$$
for the corresponding $x$ and $y$ in $G_{1}$. Then
$$
\varphi^{-1}(a) *{1} \varphi^{-1}(b)=\varphi^{-1}(\varphi(x)) *{1} \varphi^{-1}(\varphi(y))=x *_{1} y
$$
Applying bijection $\varphi$ to the right hand side we would get
$$
\varphi\left(x *{1} y\right)=a *{2} b
$$
since $\varphi$ respects multiplication. Thus $x * 1 y=\varphi^{-1}\left(a *_{2} b\right)$, so we conclude
$$
\varphi^{-1}(a) *{1} \varphi^{-1}(b)=\varphi^{-1}\left(a *{2} b\right)
$$
proving that $\varphi^{-1}$ also respects multiplication.
- The exponential function defines an isomorphism $\left(\mathbb{R}^{*}, \cdot\right) \rightarrow(\mathbb{R},+)$.
- We have encountered at least two groups of order two, namely $\left(\mathbb{Z}_{2},+\right)$ and $({1,-1}, \cdot) .$ The map
$$
\mathbb{Z}_{2} \rightarrow{1,-1}
$$
sending 0 to 1 and 1 to $-1$ gives an isomorphism between the two.
Two groups $G$ and $G^{\prime}$ are said to be isomorphic if there exists an isomorphism $\varphi: G \rightarrow G^{\prime} .$
Isomorphic groups have exactly the same properties (same order, isomorphic cyclic subgroups etc.), so we often can identify them with each other.
Claim. Being isomorphic is an equivalence relation on the set of all groups.
The main problem of the group theory is classification of groups up to the isomorphism equivalence relation.
A cyclic group of infinite order is isomorphic to $\mathbb{Z}$.
Let $n \geqslant 2$ be an integer. Any cyclic group of order $n$ is isomorphic to $\mathbb{Z}_{n}$.
8. Example 2: Direct product $\mathbb{Z}{2} \times \mathbb{Z}{2}$
The simplest example of this construction is the group $\mathbb{Z}{2} \times \mathbb{Z}{2}$ with respect to addition in both factors. It has elements
$$
{(0,0),(0,1),(1,0),(1,1)}
$$
and the operation on the above pairs via the coordinate-wise addition modulo 2 .
Let us now give some examples of how to prove that two groups are not isomorphic.
- The groups $\mathbb{Z}{4}$ and $\mathbb{Z}{2} \times \mathbb{Z}_{2}$ are not isomorphic the first one being cyclic of order 4 , whereas the second one has only elements of order at most $2 .$
- The groups $\mathbb{Q}$ and $\mathbb{Z}$ are not isomorphic. Indeed, assume we have an isomorphism $\varphi: \mathbb{Z} \rightarrow \mathbb{Q}$ and denote $\varphi(1)=a$. By surjectivity of $\varphi$, there exists an integer $n$ such that $\varphi(n)=\frac{a}{2}$. But since $\varphi$ is a homomorphism, we must have $\varphi(2 n)=2 \varphi(n)=a$, so that by injectivity of $\varphi, 2 n=1$, which is a contradiction. Note that here the argument relied on the fact that in the group $\mathbb{Q}$ one can divide by 2 indefinitely, whereas this is not possible in $\mathbb{Z}$.
Another way of seeing this is by remarking that for all $n \in \mathbb{Z}$, we have $\varphi(n)=n \varphi(1)$. This means that the denominator of the rational number $n \varphi(1)$ is at most the denominator of $\varphi(1) .$ Since the denominators of elements of $\mathbb{Q}$ can be arbitrarily large, this means that $\varphi$ cannot be surjective.
- The additive group $(\mathbb{Q},+)$ is not isomorphic to the multiplicative group $\left(\mathbb{Q}^{\times}, \cdot\right) .$ Indeed, let $\varphi$ : $\left(\mathbb{Q}^{\times}, \cdot\right) \rightarrow(\mathbb{Q},+)$ be an isomorphism. Put $\varphi(2)=a .$ By surjectivity of $\varphi$, there is a rational number $x$ such that $\varphi(x)=\frac{a}{2}$. Then $\varphi(x \cdot x)=\varphi(x)+\varphi(x)=a$, so by injectivity, $x^{2}=2$. This is impossible since there is no rational number $x$ satisfying this. This argument is similar to the one in the previous example: here we used that dividing by 2 in the additive setting corresponded to taking square roots in the multiplicative setting, which is not always possible in the rationals.
10. Cosets and Lagrange’s theorem
Let $G$ be a group and $H$ a subgroup of $G .$ A left coset of $H$ is a subset of $G$ of the form
$$
g H={g h, h \in H} \text {. }
$$
In the same way, we can define right cosets to be $H g={h g, h \in H}$ for $g \in G$.
The group $G$ itself is both its a left coset and a right coset, for $g=e$ the identity element of $G$ : $G=e G=$ Ge. More generally, for all $g \in G$, we have $G=g G=G g$.
Left and right cosets of $H$ are the same if the group $G$ is abelian, but in general they may be different. For an abelian group, we will often use additive notation and write both types of cosets in the form $g+H .$
The cosets of $H={0,3}$ in $\mathbb{Z}_{6}$ are
$$
\begin{aligned}
&0+H=3+H={0,3} \
&1+H=4+H={1,4} \
&2+H=5+H={2,5}
\end{aligned}
$$