# Show that a group of order $5$ is always abelian.

Show that a group of order $5$ is always abelian.

I know that if the binary operation $*$ is defined on a group $G$ is commutative (i.e. $a*b=b*a\ \forall a,b\in G$),\ then G is called a commutative(or abelian) group. But I don't know how to solve a group of order $5$ is abelian. Thanks in advance!

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How much group theory do you know? If you know Lagrange's Theorem, this is pretty easy. If you are only beginning to learn about groups, then this can be rather tedious. –  Arturo Magidin Apr 12 '12 at 3:04
i know some result of group theory and i know that every group of prime order is cyclic and every cyclic group is abelian that's why group of order $5$ is abelian. And this things only give me validity of given statement but i want to prove this by definition of abelian group and order of an element of a group –  Kns Apr 12 '12 at 3:44
Sorry, but that makes no sense to me. If you know that every group of prime order is cyclic, and you know every cyclic group is abelian, then why do you think this is not a proof? The implication that "every cyclic group is abelian" uses the "definition of abelian group". –  Arturo Magidin Apr 12 '12 at 16:54

Hint: Take some non-identity element $a\in G$. Show that the set $\{a^n : n\in\mathbb N\}$ is a subgroup of $G$, and that it has $5$ elements so is the whole group $G$. Thus every element of $G$ is a power of $a$, so it is easy to show that $G$ is abelian.

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You say in comments that "you know" that every group of prime order is cyclic, and that every cyclic group is abelian. I do not understand how this can give you "only" the validity, but not a proof. What's "only" about it?

If you already know that a group $G$ is cyclic, then you can prove explicitly that it is abelian as follows: $G$ is cyclic means that there exists $g\in G$ such that for every $x\in G$ there exists (at least one) $n\in\mathbb{Z}$ such that $x=g^n$.

So if $x,y\in G$, then there exist $n,m\in\mathbb{Z}$ such that $x=g^n$ and $y=g^m$. Then $xy = g^ng^m = g^{n+m} = g^{m+n} = g^mg^n = yx$, which proves that $xy=yx$ for all $x,y\in G$, hence $G$ is abelian.

If you already know that every group of prime order is cyclic, then a group of order $5$ is cyclic, and by the argument above must be abelian.

If you "know" that every group of prime order is cyclic but you don't know how to prove it (in which case I would not describe it as something you "know", but that's just me), then it becomes a little trickier. I will assume you don't know Lagrange's Theorem:

Lagrange's Theorem. If $G$ is a finite group of order $n$, and $H$ is a subgroup of order $d$, then $d$ divides $n$.

A corollary is: if $G$ is a finite group of order $n$, and $g\in G$, then the order of $g$ is a divisor of $n$.

Using Lagrange's Theorem, we have: let $G$ be a group of order $5$. Let $x\in G$, $x\neq 1$ (the identity of the group). Then the order of $x$ must divide $|G|=5$, hence $|x|=1$ or $|x|=5$. If $|x|=5$, then the cyclic subgroup generated by $x$ equals all of $G$, so $G$ is cyclic (and hence abelian as above). If $|x|=1$, then $x=1$, which we explicitly discounted. So $G=\langle x\rangle$, and we are done.

We can prove Lagrange's Theorem as follows: let $G$ be a finite group, and let $H$ be a subgroup. For every $y\in G$, let $Hy$ be the subset of $G$ given by: $$Hy = \{hy\mid h\in H\}.$$

Claim: Let $y$ and $z$ be elements of $G$. If $Hy\cap Hz\neq\varnothing$, then $Hy=Hz$.

Proof. Let $g\in Hy\cap Hz$. Then there exist $h_1,h_2\in H$ such that $g=h_1y=h_2z$. In particular, $zy^{-1}=h_2^{-1}h_1\in H$, and $yz^{-1}=h_1^{-1}h_2\in H$. We prove that $Hy=Hz$ by double inclusion.

Let $g\in Hz$. Then $g=hz$ for some $h\in H$. Note that $hh_1^{-1}h_2\in H$, so $(hh_2^{-1}h_1)y\in Hy$. But $$(hh_2^{-1}h_1)y = h(h_2^{-1}h_1)y = h(zy^{-1})y = hz(y^{-1}y) = hz= g,$$ so $g\in Hy$. Thus, $Hz\subseteq Hy$. Symmetrically, if $g=hy\in Hy$, then $$g = hy = hyz^{-1}z = h(h_1^{-1}h_2)z = (hh_1^{-1}h_2)z\in Hz,$$ so $Hy\subseteq Hz$. Thus, if $Hy\cap Hz\neq\varnothing$, then $Hy=Hz$.

Claim 2. For all $y\in G$, $y\in Hy$.

Proof. We can write $y=1y$, with $y\in H$.

Claim 3. The sets $Hy$, $y\in G$, are a partition of $G$.

Proof. Their union is everything, since $y\in Hy$ for all $y\in G$; and any two distinct sets are disjoint by Claim 1. So they form a partition.

Claim 4. If $y\in G$, then $|Hy|=|H|$.

Proof. Define $f\colon H\to Hy$ by $f(h) = hy$. This is one-to-one, since $hy=h'y$ implies, by cancellation, $h=h'$; and onto, since every element of $Hy$ is of the form $hy$ for some $h\in H$. Thus, $f$ is a bijection.

Proof of Lagrange's Theorem. Let $|G|=n$ and $|H|=d$. Let $g_1,\ldots,g_k$ be elements of $G$ such that $Hg_i\cap H{g_j}=\varnothing$ if $i\neq j$, and with $G=\cup_{i=1}^k Hg_i$. They exist since the $Hg$ form a partition of $G$ and $G$ is finite. Then $$n=|G| = \left|\cup_{i=1}^k Hg_i\right| = \sum_{i=1}^k|Hg_i| = \sum_{i=1}^k|H| = k|H| = kd,$$ so $d$ divides $n$ as desired. $\Box$

And now you have all the ingredients.

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