Every group of order 5 is abelian

Today I was reading some basic group theory from Herstein’s Topics in Algebra, and saw the following cute problem: Prove that every group of order 5 is abelian. Now, clearly Lagrange’s theorem implies that there is only one group of order 5, the cyclic group of order 5, which is obviously abelian. It would be too much cheating to resort to a such theorem, because Lagrange’s theorem is introduced later in the book. This gives me the impression that Herstein wants the reader to solve this problem by using only the definition of a group, and its basic properties (i.e. uniqueness of inverse element and etc.). I must admit I had a lot of fun coming up with a solution, but I feel like this is not the intended solution, because it is far from being an elegant argument.

Proposition. Every group of order 5 is abelian.

Proof: Assume a group G has 5 elements. To be more concrete, let G=\{a,b,c,d,e\} where e is the identity element of G. We will assume, to the contrary, that G is not abelian. Then, there exists two elements x,y\in G such that xy\ne yx. But clearly, for this statement to be true, x\ne e and y\ne e. Thus, we will assume, without loss of generality that x=a and y=b. In other words, we assume ab\ne ba.

Now, observe that ab\ne a, ba\ne a (because both implies b=e, which is false). Similarly, ab\ne b and ba\ne b (because both implies a=e which is false). Also, we claim that ab\ne e. Because otherwise, a would be inverse of b which would imply that ab=e=ba, contrary to our assumption that ab\ne ba. Similarly, ba\ne e. Thus, ab,ba\in \{c,d\}. Without of loss of generality, we will assume ab=c and ba=d.

The next step is to observe that db=(ba)b=b(ab)=bc and ad=a(ba)=(ab)a=ca. We have shown that db=bc and ad=ca. The cancellation law shows that db=bc\ne b,c,d. Also, we claim that db=bc\ne e, because otherwise we would have two elements c and d that serve as inverse for element b. But then, by uniqueness of an inverse element, it follows that c=d, contradiction! This forces db=bc=a. Similar analysis shows ad=ca\ne a,c,d,e, thus ad=ca=b.

Let’s see what we have. bc=a implies bab=a. Multiply both sides by a from the right to get (ba)^2=a^2. Similarly, ca=b implies aba=b. Multiply both sides by b from the left to get (ba)^2=b^2. This implies a^2=b^2 (because both are equal to (ba)^2).

Finally, we can analyze a^2=b^2. Clearly, a^2=b^2\ne a, b (because it would otherwise imply a=e or b=e). Furthermore, a^2\ne c because otherwise, a^2=ab, or a=b which is false. Similarly, a^2\ne d because otherwise a^2=ba, or a=b which is false. This leaves us no other choice but a^2=b^2=e. But then, cb=(ab)b=ab^2=a. But recall that db=bc=a. In particular, db=a. Combining this with cb=a, we get db=cb. Cancelling b from the right, we get d=c, which is a contradiction! This shows that having ab\ne ba is impossible, and so G must be abelian, completing the proof. \blacksquare

Advertisements
This entry was posted in Uncategorized and tagged . Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s