If group $G$ has 5 elements then $G$ abelian

Question

If $G$ - group and $|G|=5$ then $G$ must be abelian group.

My efforts: Let $G=\{a,b,c,d,e\}$ then $ab,\ ba\in G$. But it is easy to verify that $ab \neq a, b$ and if $ab=e$ then $ba=e$. Suppose that $ab=c$ then $ba\in \{c,d,e\}$

If $ba=c$ then we get $ab=ba$
If $ba=e$ then $ab=e=c$ which is conreadiction.
We have the last case $ab=c$ and $ba=d$. Let's consider $aG=\{a^2,ab,a^2b,aba,a\}$ and $Ga=\{a^2,ba,aba, ba^2,a\}$. Since $c\neq d$ then $ab\neq ba$ so $ab=ba^2$ and $a^2b=ba$. But I am not able to derive that $ab=ba$.

Can anyone explain how to continue my reasoning?

Do you need to derive it from first principles, or can you use theorems such as 'order of a subgroup divides the order of group'? — , Nov 19 '17 at 16:31
@user8734617, yes i should derive it from first principles. We have no in possess Lagrange's theorem. — RFZ, Nov 19 '17 at 16:36
Since you have asked this question, one might expect that you know already some basics in group theory:) — Dietrich Burde, Nov 19 '17 at 16:45
Which of the elements in $aG$ and $Ga$ would be the identity? What might $a^2$ be? — Mark Bennet, Nov 19 '17 at 16:49

score 3 · Accepted Answer · answered Nov 19 '17 at 16:46

3

From $ab=ba^2$ and $a^2b=ba$ you get $a^2ba=ba^2=ab$, hence $aba=b$. The identifications

$$\{a,b,c,d,e\}=\{a,b,ab,ba,e\}=G=aG=\{a^2,ab,a^2b,aba,a\}=\{a^2,c,d,aba,a\}=\{a^2,c,d,b,a\}$$

now tell us $a^2=e$. But then $d=a^2b=eb=b$, a contradiction.

answered Nov 19 '17 at 16:46

Barry Cipra

79,832

Indeed nice solution. My efforst also were too close but I have no idea to make identifications. – RFZ Nov 19 '17 at 17:57

score 1 · Answer 2 · answered Nov 19 '17 at 16:34

1

The order $o(g)$ of any element divides $5$ and as this is prime, $o(g) = 1$ and so $g=1$, or $o(g) = 5$ and $g$ generates the group: $G = \{g, g^2, g^3, g^4, 1\}$. Cyclic goups are Abelian as $g^n g^m =g^{n+m} = g^m g^n$ for all $n,m$.

answered Nov 19 '17 at 16:34

Henno Brandsma

242,131

There is no notion of cyclic groups, Lagrange's theorem in my book yet – RFZ Nov 19 '17 at 16:38
@RFZ it's easy to see from first principles. – Henno Brandsma Nov 19 '17 at 16:40
Could you show it please? I have done many ways but no results. – RFZ Nov 19 '17 at 16:40

score 0 · Answer 3 · answered Nov 19 '17 at 16:31

0

But our $G$ is just a cyclic group.

Take $a\in G$ such that $a\neq e$ and consider $\{a^n|n\in\mathbb N\}$.

You'll get $a^5=e$ and $<a>=G$.

answered Nov 19 '17 at 16:31

Michael Rozenberg

194,933

There is no notion of cyclic group yet in my book – RFZ Nov 19 '17 at 16:37
@DietrichBurde, Maybe you are right but I am going to learn this concept immediately after that set of problems – RFZ Nov 19 '17 at 16:41
@RFZ You can get it step by step. If $a^k=e$ and $1<k\leq5$ so $5$ is divided by $k$. Is it clear? – Michael Rozenberg Nov 19 '17 at 16:44

score 0 · Answer 4 · answered Nov 19 '17 at 16:57

Since every group of order $5$ is cyclic, you can draw the Cayley table as follows: \begin{array}{ c|cc } \cdot & e & a & a^2 & a^3 & a^4\\ \hline e & e & a & a^2 & a^3 & a^4\\ a & a & a^2 & a^3 & a^4 & e\\ a^2 & a^2 & a^3 & a^4 & e & a\\ a^3 & a^3 & a^4 & e & a & a^2\\ a^4 & a^4 & e & a & a^2 & a^3\\ \end{array} where $G=\{a,a^2,a^3,a^4,a^5=e\}=\langle a\rangle$.

score 0 · Answer 5 · answered Nov 19 '17 at 18:31

Suppose any element $a,b,c,d$ has order $5$, without loss of generality, say $a$.

Then $G = \{e,a,a^2,a^3,a^4\}$, and is clearly abelian (since powers commute with themselves:

$a^ka^m = a^{k+m} = a^{m+k} = a^ma^k$).

So we want to investigate what other orders (besides $5$) elements might have.

Suppose an element (again, say $a$) has order $4$, so that:

$G = \{e,a,a^2,a^3,b\}$.

It is immediate that $ab \neq e,a,b$. Which leaves us with $ab = a^2$ or $ab = a^3$.

$ab = a^2 \implies a^3b = e \implies b = a$, contradiction.

$ab = a^3 \implies a^2b = e \implies b = a^2$, contradiction.

We conclude there is no element of order $4$.

Thus we could only have elements of order $2$ or $3$. Since elements of order $3$ occur in inverse-pairs (prove this!), we either have $0,2$ or $4$ such elements, and thus correspondingly, $4,2$ or no elements of order $2$.

If we have no elements of order $3$, and thus $4$ elements of order $2$, then:

$G = \{e,a,b,ab,c\}$, with $a^2 = b^2 = (ab)^2 = c^2 = e$.

So $e = (ab)^2 = abab \implies ba = ab$, and $H = \{e,a,b,ab\}$ forms a subgroup of $G$ (why?). In this case, we have $ac \neq e,a,c$, so $ac = ab \implies c = b$, or $ac = b \implies c = ba = ab$, both of which are impossible. So not every non-identity element is order $2$.

So we assume we have (exactly) two elements of order $3$, so that:

$G = \{e,a,a^2,b,c\}$ with $a^3 = b^2 = c^2 = e$.

Again, it is immediate that $ac \neq e,a,c$. If $ac = a^2$, then $c = a$, contradiction.

So we are down to $G = \{e,a,a^2,ac,c\}$ with $a^3 = (ac)^2 = c^2 = e$.

It is clear that $a^2c \neq e,a^2,c$. If $a^2c = a$, then $ac = e$, and if $a^2c = ac$ then $a^2 = a$, both impossible. This eliminates the possibility that we have just two elements of order $3$.

Finally, we assume $G = \{e,a,a^2,b,b^2\}$ with $a^3 = b^3 = e$.

Again, it is clear that $ab \neq e,a,b$. I leave it to you to show $ab = a^2$, and $ab = b^2$ cannot be.

I freely admit there may be shorter ways to do this.

score 0 · Answer 6 · answered Nov 19 '17 at 21:20

It seems to me that a finite group of prime order is necessarily a cyclic group (hence it is abelian). Just take the sub-group generated by an element g not= id and use Lagrange’s theorem which says the order of a sub-group must divide the order of the group.

If group $G$ has 5 elements then $G$ abelian

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