Open loop gain and Percent Overshoot Relationship
"Higher the loop gain of the system, larger is the percent overshoot"
  
  Can anyone show that analytically? Consider it to be a second order
  system.


My answer is based on assuming that a 2nd order system is modified with gain and put inside a control loop hence, the system can then be regarded as having “loop gain”. This then makes it a “control-system” as per the tag in the question and, the term “loop gain” then makes sense. Other answers may not have made this assumption: -



Starting with an open-loop, 2nd order, low pass system; it may have an output/input equation similar to this: -

$$H(s) = \dfrac{\omega_o^2}{s^2 + 2\zeta\omega_o s + \omega_o^2}$$

Where \$\omega_o\$ is the natural resonant frequency and \$\zeta\$ is the damping ratio.

This is modified to include a gain factor (\$G\$) and, inserted inside a unity gain closed loop with negative feedback. The new equation for H(s) becomes: -

$$H(s) = \dfrac{\dfrac{G\cdot\omega_o^2}{s^2 + 2\zeta\omega_o s + \omega_o^2}}{1 + \dfrac{G\cdot\omega_o^2}{s^2 + 2\zeta\omega_o s + \omega_o^2}}$$

$$= \dfrac{G\cdot\omega_o^2}{s^2 + 2\zeta\omega_o s + \omega_o^2 + G\cdot\omega_o^2}$$

$$= \dfrac{G\cdot\omega_o^2}{s^2 + 2\zeta\omega_o s + \omega_o^2 (1 + G)}$$

Notable is the \$1+G\$ term in the denominator because it modifies the old natural resonant frequency from \$\omega_o\$ to a new resonant frequency (\$\omega_n\$) that is \$\sqrt{1+G}\$ times bigger.

And, the important thing here is that this has the effect of lowering the damping ratio (in other words the old damping ratio is greater than the new damping ratio i.e.\$\zeta_o &gt; \zeta_n\$). Hence, the terms in front of s (namely \$2\zeta\omega_o\$) relative to the new natural resonant frequency (\$\omega_n\$),  become: -

$$\dfrac{2\zeta\omega_o}{\omega_n} \Rightarrow\dfrac{2\zeta\omega_o}{\omega_o\sqrt{1+G}} = \dfrac{2\zeta}{\sqrt{1+G}}$$

This, in effect, lowers the old damping ratio and creates more overshoot.