Exponentiation Problem of G2 in MNT curve I made a simple python program in the Charm framework (https://github.com/JHUISI/charm):
from charm.toolbox.pairinggroup import PairingGroup,ZR,G1,G2,GT,pair

group = PairingGroup(&quot;MNT224&quot;)
g = group.random(G1)
h = group.random(G2)
a, b = group.random(ZR, 2)

G1 = (g ** a) ** b
G2 = g ** (a * b)
H1 = (h ** a) ** b
H2 = h ** (a * b)

print('G1:', G1)
print('G2:', G2)
print('H1:', H1)
print('H2:', H2)

And I got this:
G1: [10078373295334768694563741838278622956778134163588850571046784075738, 11285206652127590502238684314907500254559622599368025831501761600958]
G2: [10078373295334768694563741838278622956778134163588850571046784075738, 11285206652127590502238684314907500254559622599368025831501761600958]
H1: [[3534196784717504462328757211224470588231695063171615945530129056656, 10211780109365030693121484817970351226593726986080067998578894245699, 6368599583350992357625322855407611464986270273403678555426952468683], [13684062230274089937149579621207233764284608067400053114106045228780, 9213593495915716016992145616323404591735310637893710560096176472666, 8444322233041608777047377989157187452524603129738981272127139119302]]
H2: [[1419904059378831003656260139736181866360018467588549133311449391908, 8009525015121730906488018517095803865211202805317619827555115979123, 13352161910105351377726749012042030874411151465492265223612708819089], [13998900821711144185326068030921609854586336394019796349518302780691, 4263873462088002148509836008014306558952369669692605762808882317126, 714307621193276272340511827939733717827940292614445171003103716515]]

So why H1 is not equal to H2?
Here's a guess: when you ask to compute h ** (a * b), it actually computes h ** (a * b mod q), where $q$ is the order of $g$ in $G1$.
If that is the case, well, a random element from $G2$ is not going to be a divisor of $q$ - any element that is the result of a pairing operation will be; however most elements of $G2$ cannot be the result of a pairing operation.
And, if the order of $h$ is not a divisor of $q$, then $h^{a \cdot b} \ne h^{a \cdot b \bmod q}$
The above would give the expected result if you set $h$ to be the result of a pairing operation, possibly with two random $G1$ elements...
(This is labeled a guess because I'm not familiar with the Charm framework, so I'm guessing about its behavior)