Two groups can look different on paper and be the same group in disguise: $U(10)$ and $\mathbb{Z}_{4}$ have the same multiplication pattern once you rename the elements. An isomorphism is such a renaming.
Definition. An isomorphism is a homomorphism that is a bijection. If one exists, $G$ and $H$ are isomorphic, $G \cong H$. The inverse of an isomorphism is again a homomorphism, so $\cong$ is an equivalence relation. By the kernel test, a homomorphism is one-to-one exactly when its kernel is trivial.
Showing two groups are isomorphic
Write down a map and check the two conditions: it is a homomorphism, and it is a bijection (for finite groups of the same order, one-to-one is enough, and that is the kernel test).
Why it works
Proof. Let $G = \langle a \rangle$ with $|a| = n$, and define $\psi: \mathbb{Z}_{n} \to G$ by $\psi(k) = a^{k}$. It is well defined, because $a^{k} = a^{j}$ exactly when $k \equiv j \pmod n$ (the order lemma). It is a homomorphism: $\psi(j + k) = a^{j+k} = a^{j}a^{k}$. Its kernel is $\{k : a^{k} = e\} = \{0\}$, so it is one-to-one, and since both groups have $n$ elements it is onto. $\square$ The same argument shows every infinite cyclic group is $\cong \mathbb{Z}$.
Worked example: $U(10) \cong \mathbb{Z}_{4}$
- $U(10) = \{1, 3, 7, 9\}$ is cyclic, generated by $3$: $3^{1} = 3$, $3^{2} = 9$, $3^{3} = 7$, $3^{4} = 1$.
- Send $k \mapsto 3^{k} \bmod 10$: $0 \mapsto 1$, $1 \mapsto 3$, $2 \mapsto 9$, $3 \mapsto 7$.
- It is a bijection, and $3^{j + k} = 3^{j}3^{k}$ turns addition mod 4 into multiplication mod 10.
Answer: $k \mapsto 3^{k}$.
Showing two groups are not isomorphic
Find a property that isomorphisms preserve and that one group has and the other lacks. Isomorphisms preserve the order of the group, the order of each element (so the number of elements of each order), being abelian, being cyclic, and the size of the center. Trying one map and watching it fail proves nothing: some other map might work.
| Group (order 8) | Abelian? | Elements of order 1, 2, 4, 8 |
|---|---|---|
| $\mathbb{Z}_{8}$ | yes | 1, 1, 2, 4 |
| $U(15)$ | yes | 1, 3, 4, 0 |
| $U(24)$ | yes | 1, 7, 0, 0 |
| $D_{4}$ | no | 1, 5, 2, 0 |
| $Q_{8}$ (quaternions $\pm 1, \pm i, \pm j, \pm k$) | no | 1, 1, 6, 0 |
Worked example: Telling groups apart
- Orders in $U(15) = \{1, 2, 4, 7, 8, 11, 13, 14\}$: $|2| = |7| = |8| = |13| = 4$ and $|4| = |11| = |14| = 2$. No element of order $8$, so $U(15)$ is not cyclic, while $\mathbb{Z}_{8}$ is.
- $D_{4}$ has five elements of order 2 (four reflections and $r^{2}$); $Q_{8}$ has one ($-1$).
Answer: Different counts of elements by order.
Common mistake
Key idea