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I am trying to show that there exist only $2$ non-isomorphic groups of order $4$. I found the groups using Cayley Tables, (I think one is called the Klein group that I found, and the other one is a cyclic group generated that is isomorphic to $\Bbb Z_4$.)

To show that the two groups are not isomorphic to each other, I showed that one has $x^2=e$ for all $4$ elements, while the other one only has $2$. So, one has $|m| = 4$, and other has $|n|=2$. Thus, they are not isomorphic to each other.

I know there are many questions like this such as: Prove that there only two groups of order 4 up to isomorphism

I have found the $2$ groups already through trial and error and I also made their Cayley tables.

But, I want to prove that every other group of order $4$ is going to be isomorphic to these two. How do I show that every other group is isomorphic to them? I know that for example, $\Bbb Z_4$ is, but there are many possible groups, and listing out examples is not enough to validate that "All other order $4$ groups are isomorphic to these two."

ahulpke
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eddie
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3 Answers3

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Let $G$ be a group of order four. Then $G$ is abelian.

Since $2\mid 4$, by Cauchy's Theorem, there exists an element $g\in G$ of order two. There exist three other elements of $G$, one of which is the identity $e$; call the other two $a,b$. If $ag=e$, then $a=ae=ag^2=(ag)g=g$, a contradiction, so either: $ag=a$, which implies $g=e$, a contradiction; $ag=g$, which implies $a=e$, a contradiction; or $ag=b$. So let $ag=b$.

The orders of $a,b$ can be either two or four by Lagrange's Theorem.

Suppose $a$ has order two. Then $g=eg=a^2g=a(ag)=ab$. We have $e=g^2=(ab)^2=a^2b^2=eb^2=b^2$, so, since $b\neq e$, the order of $b$ is two. Thus $G$ must be isomorphic to $\Bbb Z_2^2$.

Now suppose $a$ has order four. Then $e, a, a^2, a^3$ are all distinct. We have either $a^2=g$ or $a^3=g$. If the latter, then $e=a^3a=ga=b$, a contradiction; hence $a^2=g$. But that leaves $a^3=b$. Thus $G$ must be isomorphic to $\Bbb Z_4$ (since it is generated by $a$).

Shaun
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  • why does imply that G is abelian at the start? – eddie Feb 11 '22 at 18:26
  • Follow the link, @eddie. – Shaun Feb 11 '22 at 18:28
  • is there a way to show this without using the Lagrange theorem or Cauchy theorem? I'm trying to do this only using the cayley table and basic ideas involving isomorphisms, cyclic groups, abelian groups, and cayley tables. I can't use the theorems because they haven't been covered yet. Is there a less advanced way to prove this? – eddie Feb 11 '22 at 18:35
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The abelianess without any appeal to orders of elements, Cauchy's Theorem or Lagrange's or type of group: assume that $G$ has $4$ elements and is not abelian. Then we can find two non-identity elements $a,b$ that do not commute, so $ab \neq ba$. Note that this implies $ab \notin \{e,a,b\}$. Because $G$ is closed under the group operation we must have $G=\{e,a,b,ab\}$. Now $ba$ belongs to this set. Since $a$ and $b$ do not commute $ba \notin \{e,a,b\}$. Hence we must have $ab=ba$, a contradiction. So $G$ is abelian after all.

OK, so we have $\{e,a,b,ab\}$ is an abelian group of $4$ elements, so each of them does not equal one of the others. So were is $a^2$? Assume for the moment that $a^2 \neq e$. Can $a^2=a$? No, since then it would follow $a=e$. Can it be $a^2=b$? Yes, and it that case $ab=a^3$ and the group looks like $\{e,a,a^2,a^3\}$. Observe that $a^4$ must be the identity since if $a^4=a$, $a^4=a^2$ or $a=a^3$, the set reduces to less than $4$ elements. So the group is cyclic of order $4$ generated by $a$, in this case.

Similarly, if $b^2=e$, we would arrive at the group $\{e,b,b^2,b^3\}$, which is of course again cyclic of order $4$ and isomorphic to the one we already found.

We are left with the case where $a^2=e=b^2$. But then (using abelianess) $(ab)^2=abab=aabb=a^2b^2=e.e=e$ and now all elements of the group have order $2$. This one is isomorphic to $C_2 \times C_2$, a direct product of two groups of order $2$, also called the Klein $4$-group $V_4$.

So you see in this small case everything can still be figured out without using more sophisticated theorems.

Nicky Hekster
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  • so are all abelian groups isomoprhic to each other in order $4$ groups? – eddie Feb 11 '22 at 18:57
  • No this only proves that a group of order $4$ is abelian. You can follow the same train of reasoning to arrive at a cyclic group or one where every non-identity element has order $2$. – Nicky Hekster Feb 11 '22 at 19:36
  • is the way I showed that the two groups I found are non isomorphic sufficient enough? – eddie Feb 11 '22 at 20:33
  • Yes you are in the right direction, but you need to make it a bit more precise. I will add a solution to the answer I already gave. Will do that over the weekend. Again no need of Cauchy or Lagrange. – Nicky Hekster Feb 11 '22 at 23:05
  • As promised I added the rest of the reasoning. See also this post https://math.stackexchange.com/questions/909733/a-group-with-five-elements-is-abelian dealing with groups of order $5$. Bit more complicated but similar reasoning. This one I used to show to my students. – Nicky Hekster Feb 12 '22 at 08:40
  • Thank you so much! I've been stuck on this problem for so long, but I think I get it now. Thank you – eddie Feb 12 '22 at 17:42
  • You are welcome! – Nicky Hekster Feb 12 '22 at 19:25
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It seems to me that you can also reason this way. Let $G$ be a group of order 4.

  1. Every element of $G$ other than $e$ is of order 2 or 4. In fact, if $a\in G$, then $|a|\leq 4$. Let $|a|=3$, and $b\notin\{e,a,a^2\}$. It is easy to see that $ab\notin\{e,a,a^2\}$ but then $|G|>4$. Contradiction.

  2. If $G$ has an element of order $4$, then $G$ is a cyclic group.

  3. If all elements of $G$ except $e$ are of order $2$, then the group $G$ is abelian ($ab=a^{-1}b^{-1}=(ba)^{-1}=ba$). Let $a$ and $b$ be two different non-identity elements of $G$. We have $G=\{e,a,b,ab\}$. This means that any two such groups are isomorphic.

It follows that every group of order 4 is either a cyclic group or the Klein four-group.

kabenyuk
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