A finite abelian group whose order is divisible by a prime $p$ contains an element of order $p$ The following proof is from "An Introduction to theory of groups" by J . J Rotman.

Theorem-If $G$ is a finite abelian group whose order is divisible by prime $p$, then $G$ contains an element of order $p$.

Proof:
Write $|G| = pm$, where $m \ge$ 1. We proceed by induction on $m$ after
noting that the base step is clearly true. For the inductive step, choose $ x\in  G$
of order $t > 1$. If $\ p\mid t\ $ then $\ x^{t/p}$ is of order  $p$, and the
lemma is proved. We may, therefore, assume that the order of $x$ is not divisible
by $p$. Since $G$ is abelian, $\langle x\rangle $ is a normal subgroup of $G$, and $G/\langle x \rangle $ is an
abelian group of order $|G|/t = pm/t$. Since $p$ does not divide $t$, we must have $m/t < m$ an
integer. By induction, $G/\langle x\rangle $ contains an element $y^*$ of order $p$. But the
natural map $v:G\to G/\langle x\rangle $ is a surjection, and so there is $y\in G$ with $v(y) =
y^*$.The order of $y$ is a multiple of $p$, and we have returned to
the first case. $\square$
How is induction being used here?
I am assuming they are talking about principle of mathematical induction where we prove that a give statement is true for all natural number by proving it for $n=1$, then assuming it is true for some n=m then proving it for $n=m+1$. I understand up to the part where they say $m/t$ is an integer. But after that how are they saying by induction $G/\langle x\rangle $ contains an element of order $p$. Where is the part where we assume the statement to be true for $n=m$ then prove it for $m+1$.
 A: Here strong induction is used.

Strong induction is a variant of induction, in which we assume that the statement holds for all values preceding $k$.

Look at here.
The induction is on $m$ where $|G|= pm$ for an arbitrary abelian $G$.
In your case you have a quotient of $G$ which is abelian and has order less than $G$.
A: Here we are using strong induction. In normal induction we assume the statement is true for $ n=m$ then prove it for $n=m+1$. But in strong induction we assume that the statement is true for all $n<m$ and try to prove it for $n=m$.
If $|G|=pm$, then the statement of the theorem is assumed to be true for all $n<m$. If we take an arbitrary element $x$ from this group whose order is not a multiple of $p$, then as shown in the proof we can generate a quotient group $G/H $where $H=<x>$ which is also finite abelian and whose order is $pn$ where $n<m$. Since the theorem is true for all $n<m$. There exist an element $yH$  of order $p$ in this group where $y\in G$ and since $yH$ is an homomorphic image of y, order of y is some multiple of $p$.
