Isomorphism of groupsAQA A-Level Further Maths: Revision notes

Two groups $(G,*)$ and $(H,\circ)$ are isomorphic, written $G\cong H$, if there is a bijection $\phi:G\to H$ with $\phi(a*b)=\phi(a)\circ\phi(b)\quad\text{for all }a,b\in G.$ Such a $\phi$ is an isomorphism. It is a one-to-one pairing of the elements which also matches the operation, so the groups have the same structure under different labels. Their Cayley tables become identical when the elements are matched up. An isomorphism sends the identity to the identity, and inverses to inverses: $\phi(e_G)=e_H$ and $\phi(a^{-1})=\phi(a)^{-1}$.

Isomorphic groups have the same order, so first compare the numbers of elements. For two cyclic groups of the same order, an isomorphism is easy to write down: if $C=\langle c\rangle$ and $H=\langle h\rangle$ both have order $n$, then $\phi(c^k)=h^k$ is an isomorphism, because $\phi(c^jc^k)=h^{j+k}=h^jh^k$. Example: $H=\{0,1,2,3\}$ under addition modulo $4$ and $G=\{1,i,-1,-i\}$ under multiplication. Both are cyclic of order $4$ (generators $1$ and $i$), so $\phi(n)=i^n$ gives $0\mapsto1$, $1\mapsto i$, $2\mapsto-1$, $3\mapsto-i$. Check: $\phi(2+_43)=\phi(1)=i=(-1)(-i)=\phi(2)\phi(3)$. Example: $\{0,\dots,5\}$ under $+_6$ and $\{1,\dots,6\}$ under multiplication modulo $7$. The powers of $3$ modulo $7$ are $3,2,6,4,5,1$, so $3$ generates and $\phi(k)=3^k$ is an isomorphism.

Because $\phi(a^n)=\phi(a)^n$, an element and its image have the same order. So isomorphic groups have the same number of elements of each order. They also have the same order, the same number of subgroups of each size, and are both cyclic or both not, and both abelian or both not. What is not preserved is the labelling: the elements may be numbers, complex numbers or symmetries, and the operation may be addition or multiplication.

Find one property that an isomorphism preserves but the two groups do not share.

1. Different orders (numbers of elements).
2. Different numbers of elements of some order. Example: $\{1,3,5,7\}$ under multiplication modulo $8$ has three elements of order $2$; the cyclic group $C_4$ has one. Not isomorphic.
3. One is cyclic, the other is not. A cyclic group of order $n$ has an element of order $n$.
4. One is abelian, the other is not. The symmetry group of an equilateral triangle is not abelian, and $C_6$ is, so they are not isomorphic. State the property, say why isomorphisms preserve it, and compare the two groups.

Let $H$ have order $6$ with an element $h$ of order $6$. Then $h,h^2,\dots,h^6=e$ are six distinct elements, so $H=\langle h\rangle$ is cyclic. Define $\phi(c^k)=h^k$ from $C=\langle c\rangle$. It is a bijection, and $\phi(c^jc^k)=h^{j+k}=h^jh^k$, so $\phi$ is an isomorphism. The element $h^4$ has order $\frac{6}{\gcd(4,6)}=3$, matching $c^4$. $H$ is not isomorphic to the symmetry group of an equilateral triangle, which has no element of order $6$.