From my lecture notes I have that:
The Lie algebra is introduced as a set of matrices $\theta$ that gave rise to elements $M$ of the corresponding matrix group through $M=e^{i\theta}$. However, just like a group can be thought of as an abstract set of instead of matrices, we can also give an abstract definition of the Lie algebra. For an abstract group, the defining property of the group is the group operation, i.e., the product $M_1M_2$ of two group element s $M_1$ and $M_2$. Assuming infinitesimal transformations, we write them in terms of infinitesimal elemnts $\theta_1$ and $\theta_2$ of the Lie algebra as $M_1=e^{i\theta_1}$ and $M_2=e^{i\theta_2}$. We want to find the element of the lie algebra that corresponds to the product $M_1M_2$. We can Taylor expand the product as $$M_1M_2=\left(\Bbb{I}+i\theta_1-\frac12\theta_1^2\right)\left(\Bbb{I}+i\theta_2-\frac12\theta_2^2\right)+\mathcal{O}(\theta^3)$$ $$=\Bbb{I}+i\left(\theta_1+\theta_2\right)-\frac12\left(\theta_1^2+\theta_2^2+2\theta_1\theta_2\right)+\mathcal{O}(\theta^3)$$ $$=\Bbb{I}+i\left(\theta_1+\theta_2\right)-\frac12\left(\theta_1+\theta_2\right)^2-\frac12[\theta_1,\theta_2]+\mathcal{O}(\theta^3)$$ $$=e^{i\left(\theta_1+\theta_2+i\frac12[\theta_1,\theta_2]\right)}+\mathcal{O}(\theta^3)\tag{1}$$ This means that the element of the Lie algebra that corresponds to the product $M_1M_2$ is $\theta_1+\theta_2+i\frac12[\theta_1,\theta_2]$, which shows that $i[\theta_1,\theta_2]$ has to be an element of the Lie algebra, too. Furthermore, this shows that group product rules are determined by the commutators $[\theta_1,\theta_2]$.
In the passage above, $\Bbb{I}$ is the identity matrix.
I'm trying to understand how the last equality follows from the preceding one.
This is what I thought was going on:
Since the exponential of a matrix is the Taylor series of the exponential, if I can write the LHS as $$M_1M_2=e^{i\theta_1}e^{i\theta_2}=e^{i(\theta_1+\theta_2)}$$
then I could use Euler's formula, by writing $$e^{i(\theta_1+\theta_2)}=\cos(\theta_1+\theta_2)+i\sin(\theta_1+\theta_2)$$
and use the following trigonometric Taylor series,
$$\cos \left(\theta_1+\theta_2\right) = \sum^{\infty}_{n=0} \frac{(-1)^n}{(2n)!} \left(\theta_1+\theta_2\right)^{2n} = \Bbb{I} - \frac{\left(\theta_1+\theta_2\right)^2}{2!} + \frac{\left(\theta_1+\theta_2\right)^4}{4!} - \cdots$$
$$\sin \left(\theta_1+\theta_2\right)= \sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!} \left(\theta_1+\theta_2\right)^{2n+1} = \left(\theta_1+\theta_2\right) - \frac{\left(\theta_1+\theta_2\right)^3}{3!} + \frac{\left(\theta_1+\theta_2\right)^5}{5!} - \cdots$$
The problem is that I don't think I can use the result $e^{i\theta_1}e^{i\theta_2}=e^{i(\theta_1+\theta_2)}$ since $\theta_1$ and $\theta_2$ are matrices and I can only use that property if $M_1$ and $M_2$ commute as I read here.
So I'm completely stuck and I have no idea why $$\Bbb{I}+i\left(\theta_1+\theta_2\right)-\frac12\left(\theta_1+\theta_2\right)^2-\frac12[\theta_1,\theta_2]+\mathcal{O}(\theta^3)=e^{i\left(\theta_1+\theta_2+i\frac12[\theta_1,\theta_2]\right)}+\mathcal{O}(\theta^3)$$
Can someone please explain this to me?
Response to comment:
It is mentioned in the comment by @Callum that the second to last equality in $(1)$ is the Taylor expansion of $e^{i(\theta_1+\theta_2 +i\frac{1}{2} [\theta_1, \theta_2])}$. I'm still finding it hard to see why this is the case though, so using the Taylor series for the matrix exponential, $$e^{iX}=\sum_{n=0}\frac{\left( iX\right)^n}{n!}=\Bbb{I}+iX + \frac{i^2X^2}{2}+\cdots$$ and substituting $X=\theta_1+\theta_2 +i\frac{1}{2} [\theta_1, \theta_2]$ then, $$e^{i(\theta_1+\theta_2 +i\frac{1}{2} [\theta_1, \theta_2])}$$ $$=\Bbb{I}+i\left(\theta_1+\theta_2 +i\frac{1}{2} [\theta_1, \theta_2]\right) + \frac{i^2\left(\theta_1+\theta_2 +i\frac{1}{2} [\theta_1, \theta_2]\right)^2}{2}+\cdots$$ $$=\Bbb{I}+i\left(\theta_1+\theta_2\right)-\frac12[\theta_1,\theta_2]$$ $$-\frac12\left(\theta_1+\theta_2+i\frac{1}{2} [\theta_1, \theta_2]\right)\left(\theta_1+\theta_2+i\frac{1}{2} [\theta_1, \theta_2]\right)+\cdots$$
The first three terms of the final equality do match up with the terms in $(1)$ but the fourth term just gets messier as I multiply out the brackets and I cannot seem to show that it is equal to $-\frac12\left(\theta_1+\theta_2\right)^2$ as required by $(1)$.
I could typeset all the steps in trying to show (unsuccessfully) that $-\frac12\left(\theta_1+\theta_2+i\frac{1}{2} [\theta_1, \theta_2]\right)\left(\theta_1+\theta_2+i\frac{1}{2} [\theta_1, \theta_2]\right)=-\frac12\left(\theta_1+\theta_2\right)^2$, this would take a while but instead of doing that I feel like I am missing the point and there is some rule or manipulation I should be invoking.
To summarize, I have three out of the four terms in $(1)$ but I cannot seem to get the $-\frac12\left(\theta_1+\theta_2\right)^2$ term. May I please have some help to get this term?