# some short proofs on group theory

Let $G=\{a_2, a_2, a_3...a_n\}$ be a finite abelian group of order $n$. Let $x=a_1 a_2 a_3...a_n$. Prove that $x^2=e$.

My solution: Well since the order of the group is finite, we know the order of the elements of the group are also finite. So there exists a certain $k \in \mathbb{Z}$ such that $a^k = e$. Since the order the $n$, it must be that $k = n$. So since $x=a_1 a_2 a_3...a_n$ then $x^2 =(a_1 a_2 a_3...a_n)^2$ and because the group is abelian we have $x^2=a_1^2 a_2^2 a_3^2...a_n^2$ but $a^n = e$ and $n=2$ (DID I GO WRONG HERE??) then we must have $x^2 = eee...e = e$

Let $G$ be a group and let $a \in G$. Prove $\langle a\rangle = \langle a^{-1}\rangle$.

I don't even know how to begin this one. I know the cyclic subgroup generated by $a$ is the set $\{a^n | n \in \mathbb{Z}\}$

Show that $U_5$ is isomorphic to $U_{10}$.

Well both of these have order 4. Theorem: Every finite cyclic group of order n is isomorphic to $\mathbb{Z}$. I don't know how to apply the theorem to the units.

General question: Are all unit groups ie, $U_5$ cyclic?

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Please ask only one question per post. –  Bill Dubuque Mar 29 '11 at 3:38
Sure, what should I do with this? I grouped them together because I thought they were small and generally easy proofs. –  Tyler Hilton Mar 29 '11 at 3:41
It's okay to group questions that are related or sequential. Here, the questions are pretty separate, so they should be separate (think about someone trying to search the site to find a particular question). –  Arturo Magidin Mar 29 '11 at 3:54

Your argument in the first part is very problematical. First, it is false that "because the group is of order $n$, it must be that $k=n$." For example, in the Klein $4$-group, which has order $4$, you have $x^2 = e$ for all $x$. You can certainly conclude that $n$ works, but you cannot conclude that $n$ is the only thing that works.

Instead try this Hint: Pair each element with its inverse; if an is its own inverse, pair it with itself. Check what happens.

Question 2: In order to show that $\langle a\rangle = \langle a^{-1}\rangle$, it suffices to show that $a\in\langle a^{-1}\rangle$ and that $a^{-1}\in\langle a\rangle$. Why? Because $\langle x\rangle$ is the smallest subgroup of $G$ that contains $x$. So if $x\in H$ for any subgroup $H$ of $G$, then $\langle x\rangle\subseteq H$. So if you can prove that $a\in\langle a^{-1}\rangle$, then since $\langle a^{-1}\rangle$ is a subgroup, then you must have $\langle a\rangle\subseteq \langle a^{-1}\rangle$. Showing $a^{-1}\in\langle a\rangle$ will give the other inclusion.

Question 3: You are incorrect in your assertion that "every cyclic subgroup of order $n$ is isomorphic to $\mathbb{Z}$", but I suspect you meant $\mathbb{Z}_n$, not $\mathbb{Z}$.

There are two, and only two, groups of order $4$: the cyclic group of order $4$, and the Klein $4$-group. How can you tell them apart? Well, the cyclic group of order $4$ has an element of order $4$, but the Klein $4$-group does not. You know this group has order $4$. If all elements have order $1$ or $2$, then it's the Klein $4$-group. If there is at least one element of order $4$, then it's cyclic of order $4$.

(In general, if $G$ has order $n$, then $G$ is cyclic if and only if it has an element of order $n$).

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HINTS

1) inversion is bijection (permutation) on $\rm\:G\:,\:$ so $\rm\: G\ =\ \{a_1,a_2,\cdots,a_n\}\ =\ \{a_1^{-1},a_2^{-1},\cdots,a_n^{-1}\}\:.$ Hence taking the product of all elements both ways yields:

$\rm\quad\quad\quad x\ =\ a_1\cdots\: a_n\ =\ a_1^{-1}\cdots\: a_n^{-1}\ =\ (a_1\cdots\: a_n)^{-1}\: =\ x^{-1}\ \ \Rightarrow\ \ x^2\ =\ 1$

More generally see Wilson's theorem for groups.

2) $\rm\ \ a^{-1} \in \langle a\rangle\$ and $\rm\ a \in \langle a^{-1}\rangle$

3) $\rm\ \ p\:$ odd $\rm\:\Rightarrow\ \phi(p)\ =\ \phi(2\:p)\:,\$ i.e. $\rm\ p-1\ =\:$ the number of odd naturals below $\rm\:2\:p\:$ excluding $\rm\:p\:.$

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1. I do not see why you are making the assumption that $n = 2$, as that is only a special case of the theorem you are being asked to prove. The key here is to observe that $x$ is the product of ALL elements of $G$. For every $a_i$ there is an $a_j$ such that $a_j = a_i^{-1}$, and so if $i \neq j$ we can cancel $a_i$ and $a_j$ in $x$. We are then left with only elements that are their own inverses, and so $x^2 = e$.

2. This follows from the fact that $-1$ is a unit in $\mathbb{Z}$, and so $\{x^n | n\in \mathbb{Z}\} = \{x^{-n} | n\in \mathbb{Z}\}$.

3. Count the number of elements of $U_5$ and $U_{10}$, and check the orders of each elements. This should give you an answer.

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The group $U_5$ is such that all elements in the group are relatively prime to the congruence class modulo 5. I may have butchered the definition. But as an example $U_5 = \{1, 2, 3, 4\}$ –  Tyler Hilton Mar 29 '11 at 4:05
@Tyler: Thanks, I know that as $U(\mathbb{Z}_5)$ but I'm sure your notation is valid as well. –  Alex Becker Mar 29 '11 at 4:33

Q1. For the first question, since $G$ is a group, each element $a_i$ has an inverse $a_i^{-1}$. Note that inverses are unique. Let $\{b_i\}_{i=1}^k$ denote the subset of elements of order $2$ (I.e., $b_i = b_i^{-1}$ for all $i$). Hence,

$$x^2 = (a_1 \dots a_n)^2 = (b_1 \dots b_k)^2 = b_1^2 \dots b_k^2 = e$$

as $G$ is abelian. I apologize that this argument is not as slick as possible, but I hope it gets the point across. Feel free to accept any post that answers all three questions, by the way. I will try to type up the other two, but someone will probably beat me to it.

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Thankyou. But is it always true that if an element $a$ has order 2 then $a=a^{-1}$? –  Tyler Hilton Mar 29 '11 at 3:48
@Tyler: Yes, as the definition of $a^{-1}$ is that $aa^{-1} = a^{-1}a = e$. –  Alex Becker Mar 29 '11 at 3:50
By definition, an element $g$ of a group has order $k$ if $k$ is the smallest positive integer such that $a^k = e$. Hence, if $a$ has order $2$, then $a^2 = e$ (and $a \neq e$). Hence, $a \cdot a = e \Longrightarrow a = a^{-1}$. –  JavaMan Mar 29 '11 at 3:51
@Tyler: Yes, because if $a$ has order $2$, then $1 = a^2 = aa$, and $a^{-1}$ is the unique element $x$ such that $ax=1$. Since $x=a^{-1}$ and $x=a$ both work, then they must be equal, so $a=a^{-1}$. Conversely, if $a=a^{-1}$, then $a^2=aa=aa^{-1}=1$, so $a$ has order dividing $2$ (hence either order $2$ or order $1$). –  Arturo Magidin Mar 29 '11 at 3:51