Dgdrxrt

I am trying to verify the following proposition:

Let $G$ be a finite Abelian group and let $p$ be a prime that divides the
order of $G$. Then $G$ has an element of order $p$.

My proof:
By Lagrange's theorem: $x^G=e$. By assumption we have $kp=|G|$ for some prime $p$. So $e=x^G=x^kp$. Thus $|x^k|=p$. $blacksquare$

The book's proof uses induction and cosets -- is this necessary?

For reference, here's the book proof:

Clearly, this statement is true for the case in which $G$ has ­order 2.
We prove the theorem by using the Second Principle of Mathematical Induction on $|G|$. That is, we assume that the statement is true for all Abelian groups with fewer elements than G and use this assumption to show that the statement is true for G as well. Certainly, G has elements of prime order, for if $|x| = m$ and $m = qn$, where $q$ is prime, then $|x^n| = q$. So let $x$ be an element of $G$ of some prime order $q$, say. If $q=p$, we are finished; so assume that $q neq p$. Since every subgroup of an Abelian group is normal, we may construct the factor group $barG = G/langle xrangle$. Then $barG$ is Abelian and $p$ divides $|G|$, since $|barG| = |G|/q$. By induction, then, $G$ has an element — call it $ylangle xrangle$ — of order $p$. Then, $(ylangle xrangle)^p = y^plangle xrangle = langle xrangle$ and therefore $y^p in langle xrangle$. If $y^p = e$, we are done. If not, then $y^p$ has order $q$ and $y^q$ has order $p$. $blacksquare$

Your proof is incorrect. You know that $(x^k)^p=e$ (for any $xin G$), but this does not necessarily mean $x^k$ has order $p$. All it tells you is that the order of $x^k$ divides $p$, so it is either $1$ or $p$.

In fact, your approach cannot work without some major modification. It is possible that $x^k=e$ for all $xin G$, so there is no element of the form $x^k$ which has order $p$. For instance, if $G=(mathbbZ/pmathbbZ)^2$, then $|G|=p^2$ so $k=p$, but $x^p=e$ for all $xin G$.

From $x^kp=e$, what you can deduce is that $(x^k)^p=e$, and therefore, that $|x^k|$ divides $p$; that is, that it is equal to $p$ or equal to $1$ (that is, $x^k=e$). How do you know that it is equal to $p$?

And where did you use that $G$ is Abelian?

Here is a simple proof due to Frobenius, I believe.
Let $x_1, x_2, ..., x_n$ be the elements of the group $G$. Let $r_i$ be the order of the element $x_i$. Furthermore, let $$Z = big( mathbbZ / r_1 mathbbZ, + big) times big( mathbbZ / r_2 mathbbZ, + big) times ldots times big( mathbbZ / r_n mathbbZ, + big)$$
Now define a map $varphi$ from $Z$ into $G$ by $varphi(k_1, k_2, ldots, k_n) = x_1^k_1 x_2^k_2 dots x_n^k_n$. As $G$ is abelian it is easy to see that $varphi$ is a homomorphism. Moreover $varphi$ is surjective as $varphi(0,0, ldots, 1, ldots, 0) = x_i$. If $K = ker varphi$, then $Z/K cong G$ and therefore $|Z| = |K| cdot |G|$. Since $p$ divides $|G|$, it follows that $p$ divides $|Z|$. But $|Z| = r_1 times r_2 ldots r_n$. Hence one of the $r_i$ is divisible by $p$.