Algebra: Does every group whose order is a power of a prime p contain an element of order p? I need to solve this using Lagrange's theorem and simple corollaries. Thanks.
 A: Assuming you mean a non-trivial group, then the answer is "yes".
Let $|G|=p^n$ for some $n\ge 1$ and $x\in G$ be a non-identity element. Then $x$ has order $p^k>1$ for some $1\le k\le n$ by Lagrange. Then $x^{p^{k-1}}$ has order $p$ by definition of order.
A: Let $p^n$ be the order of a group $G$. If $n=1$, it is obvious, since the order of every element of the group $x\neq 1$ divides $p$.
Induction: Suppose that it is true for groups of order $p^k$ for $k<n$. Now, pick an element of the group $x\neq 1$.
If the order of $x$ is not $p^n$, the subgroup $H$ generated by $x$ has order less than $p^n$. Since this order has to divide $p^n$, it must be $p^k$ for some $k<n$, and by induction hypothesis, $H$ has an element of order $p$, and so has $G$.
If the order of $x$ is $p^n$, then the order of $x^p$ is clearly $p^{n-1}$. Now let $H$ be the subgroup generated by $x^p$ and reason like before.
A: Hint: if the elements of the group all have order $1$ what can you say about the group? 
Now check you can pick an element which isn't of order $1$ - what does Lagrange tell you about the possible orders for this element? 
Then think about the subgroup generated by the element you have chosen - it has an important extra structural feature which you can't guarantee in the original group - can you see how you can use the structure to find an element of order $p$.
