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I came across this post on Math Stack Exchange where OP asks about two groups which have same order but are not isomorphic.

And example given is the Klein-4 group, and $\mathbb{Z}_{4}$ among many others. But no one explained why these are not isomorphic. For something to be isomorphic, we NEED to define a function between one group, and the other right? How can we just look at groups and say one is isomorphic to the other?

Is there a hole in my intuitive understanding for what it means for two groups to be isomorphic to one another?

Shaun
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3 Answers3

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In practice, it is much easier to determine that two groups are not isomorphic than to show that they are. To show that they're not isomorphic, it suffices to find a single different property.

For example, there is no element of the Klein $4$ group that has order $4$. The largest order of an element is $2$. The cyclic group $\mathbb{Z}/4\mathbb{Z}$ has an element of order $4$. Thus these two groups are not isomorphic.

Other useful aspects that can frequently be used to show that two groups are not isomorphic include cardinality, conjugacy classes, sizes and numbers of subgroups, among others.

But it is still possible to find pairs of groups that share practically all easily computable aspects noted above, but still aren't isomorphic. This is a pretty annoying problem and very slow to compute for general finite groups. For general infinite groups, the isomorphism problem is undecidable! (M.O. Rabin, Recursive unsolvability of group theoretic problems. Ann. Math 67:172--194, 1958).

davidlowryduda
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    So, why is it the case that just because we don't have two elements with the same highest order, we can immediately disregard it as being isomorphic? –  Oct 18 '21 at 17:45
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In the Klein four group, every nontrivial element has order two, whereas $\Bbb Z_4$ has an element of order four.

For any group isomorphism $\varphi: G\to H$, we must have $\lvert \varphi (g)\rvert=\lvert g\rvert$ for all $g\in G$. The proof is routine.

Shaun
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A standard way to get non-isomorphic finite groups of equal order is the following:

Write $$G=N\rtimes_\varphi H,$$ as a product of two subgroups $N,H$ of which $N\trianglelefteq G$ is normal. The multiplication in the group $G$ is defined as $$ g_1\cdot g_2=(n_1,h_1)\cdot(n_2,h_2)=(n_1\varphi(h_1)(n_2),h_1h_2)=n_1\varphi(h_1)(n_2)h_1h_2 $$ where $\varphi :H\longrightarrow \operatorname{Aut}(N)$ is a homomorphism from $H$ to the automorphism group $\operatorname{Aut}(N)$ of $N.$ In case $\varphi =\operatorname{id}_N$ we speak of a direct product and of a semidirect product in all other cases. In case there is a group of the same order which cannot be written as a product, we have an additional example.

You can see that we get as many examples of groups of the same finite order as we have automorphisms of $N$ available.

In case of the Klein 4-group we have $G=\mathbb{Z}_2\rtimes_\varphi \mathbb{Z}_2$ and $\operatorname{Aut}(\mathbb{Z}_2)=\{\operatorname{id}_{\mathbb{Z}_2}\}\cong \{1\}$ only one automorphism, hence a direct product. However, $G=\mathbb{Z}_4$ is the other group with four elements. It cannot be written as a product of subgroups because e.g. the single element $3$ generates the entire group: $\{3,3+3=6=2,2+3=5=1,1+3=4=0\}.$

Marius S.L.
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