$H$ is a subgroup of $G$ so that $G:H=5$
$ a \in Z(G) , ord(a)=3$
Show that $ a \in H $
What I did:
if $ a \notin H $ then:
$ G/H = \{ H, aH, a^2H, gH, agH \} $ for some $ g \in G-H $
But I don't know how to procceed...
$H$ is a subgroup of $G$ so that $G:H=5$
$ a \in Z(G) , ord(a)=3$
Show that $ a \in H $
What I did:
if $ a \notin H $ then:
$ G/H = \{ H, aH, a^2H, gH, agH \} $ for some $ g \in G-H $
But I don't know how to procceed...
As in Cihan's answer, let $A = \langle a\rangle$. Since $a \in Z(G)$, $AH$ is a subgroup of $G$ containing $H$. As $[G : H]$ is prime, $AH = H$ or $AH = G$. In the latter case, $G = H\cup aH \cup a^2H$, a union of at most 3 left cosets of $H$, contradicting $[G:H] = 5$. So $AH = H$, which implies $a \in H$.
Suppose that $a\notin H$, and let $H,aH,a^2H,g_1H$, and $g_2H$ be the left cosets of $H$. Clearly left multiplication by $a$ permutes $H,aH$, and $a^2H$. It also permutes the other two cosets: $ag_1H$ must be $g_1H$ or $g_2H$, and $ag_1H=g_1aH\ne g_1H$, since $a\in Z(G)$ and $aH\ne H$, so $ag_1H=g_2H$. Similarly, $ag_2H=g_1H$. But then $g_1H=a^3g_1H=a^2g_2H=ag_1H=g_2H$, which is a contradiction.
$G$ acts on the left cosets of $H$ by left multiplication; this gives a map from $G$ to $S_5$. The action of $a$ has exponent $3$, so it is either trivial or else corresponds to a $3$-cycle. In particular, it must fix some coset, $gH$, so that $agH = gH$. This implies that $g^{-1}agH = H$, but since $a$ is central, $g^{-1}ag = a$. Therefore, $aH=H$, so $a\in H$.
This generalizes to non-prime index cases:
Proposition. Suppose $H$ is a subgroup of $G$ of index $n$, and let $p$ be a prime such that $\gcd(n,p)=1$. If $a\in Z(G)$ has order $p$, then $a\in H$.
Proof. Let $G$ act on the left cosets of $H$ by left multiplication; this induces a homomorphism $G\to S_n$, and the image of $a$ has exponent $p$. Writing the image of $a$ as a product of disjoint cycles, we see that each cycle has length either $1$ or $p$. Since $p$ does not divide $n$, the image of $a$ must have a fixed point. Therefore, there exists a coset $gH$ such that $agH = gH$, which implies $g^{-1}agH=H$. Since $a$ is central, $g^{-1}ag=a$, so we have $aH=H$, hence $a\in H$, as claimed. $\Box$
Added. As Serkan notes in a comment I only saw after posting the above, the hypotheses of the proposition can be further weakend to let the order of $a$ be $m$, where every expression of $n$ as a sum of divisors of $m$ must involve at least one summand equal to $1$.
Let $Z = Z(G)$. Since $ZH = HZ$, $ZH$ is a subgroup of $G$. We have $5 = |G:H| = |G:ZH||ZH:H|$ so either $|G:ZH| = 1$ or $|ZH:H| = 1$. In the latter case, we get $ZH = H$; hence $Z \subseteq H$ and we are done.
In the former case, $G = ZH$. So $|Z : Z \cap H| = |ZH:H| = |G:H| = 5$ Let $A = \langle a \rangle$. We know $A \subseteq Z$. Since $Z$ is an abelian group, $A(Z \cap H)$ is a subgroup of $Z$. Now we have $5 = |Z : Z \cap H| = |Z : A(Z \cap H)||A(Z \cap H) : Z \cap H|$ If $A \subseteq Z \cap H$, we are done. So suppose $A(Z \cap H) > Z \cap H$. Then by above, we get $Z = A(Z \cap H)$ and $|Z : Z \cap H| = 5$. But then $|A: A \cap Z \cap H| = |A(Z \cap H): Z \cap H| = 5$ and this contradicts $|A| = 3$.