Any compact matrix group $G$ contains a compact abelian matrix group which is (a) pathconnected and (b) nontrivial.
Optional: Finding compact abelian subgroups
Compact abelian subgroups exist
This video is optional.
In the next video, we'll see that any compact, pathconnected abelian group is a torus, so together these results will show that $G$ contains a nontrivial torus.
Let $\mathrm{\pi \x9d\x94\u20af}\beta \x8a\x82\mathrm{\pi \x9d\x94\u20ac}$ be a nontrivial abelian subalgebra, that is a subspace of $\mathrm{\pi \x9d\x94\u20ac}$ on which the commutator bracket vanishes.
Note that there is always a nontrivial abelian subalgebra if $\mathrm{\pi \x9d\x94\u20ac}$ is nontrivial, for example a line in $\mathrm{\pi \x9d\x94\u20ac}$ is a nontrivial abelian subalgebra: suppose we look at the line spanned by $X\beta \x88\x88\mathrm{\pi \x9d\x94\u20ac}$ , then any two elements ${\mathrm{\Xi \xbb}}_{1}\beta \x81\u2019X$ and ${\mathrm{\Xi \xbb}}_{2}\beta \x81\u2019X$ of this line commute with one another.
We now make two claims:

$\mathrm{exp}\beta \x81\u2018(\mathrm{\pi \x9d\x94\u20af})=\{\mathrm{exp}\beta \x81\u2018(X):X\beta \x88\x88\mathrm{\pi \x9d\x94\u20af}\}$ is an abelian subgroup of $G$ .

while $\mathrm{exp}\beta \x81\u2018(\mathrm{\pi \x9d\x94\u20af})$ is not necessarily compact, its topological closure $\stackrel{{\rm B}\u2015}{\mathrm{exp}\beta \x81\u2018(\mathrm{\pi \x9d\x94\u20af})}$ is a compact abelian subgroup.
Together, these claims will prove the proposition (we just need to take the pathcomponent of the identity in $\stackrel{{\rm B}\u2015}{\mathrm{exp}\beta \x81\u2018(\mathrm{\pi \x9d\x94\u20af})}$ ).
To prove claim 1, we observe that $$\mathrm{exp}\beta \x81\u2018({X}_{1})\beta \x81\u2019\mathrm{exp}\beta \x81\u2018({X}_{2})=\mathrm{exp}\beta \x81\u2018({X}_{1}+{X}_{2})$$ for all ${X}_{1},{X}_{2}\beta \x88\x88\mathrm{\pi \x9d\x94\u20af}$ because ${X}_{1}$ and ${X}_{2}$ commute. This implies that $\mathrm{exp}\beta \x81\u2018(\mathrm{\pi \x9d\x94\u20af})$ is a subgroup of $G$ , because if $\mathrm{exp}\beta \x81\u2018({X}_{1})\beta \x88\x88\mathrm{exp}\beta \x81\u2018(\mathrm{\pi \x9d\x94\u20af})$ and $\mathrm{exp}\beta \x81\u2018({X}_{2})\beta \x88\x88\mathrm{exp}\beta \x81\u2018(\mathrm{\pi \x9d\x94\u20af})$ then $\mathrm{exp}\beta \x81\u2018({X}_{1})\beta \x81\u2019\mathrm{exp}\beta \x81\u2018({X}_{2})=\mathrm{exp}\beta \x81\u2018({X}_{1}+{X}_{2})$ and ${X}_{1}+{X}_{2}\beta \x88\x88\mathrm{\pi \x9d\x94\u20af}$ because $\mathrm{\pi \x9d\x94\u20af}$ is a subspace.
Moreover, $\mathrm{exp}\beta \x81\u2018(\mathrm{\pi \x9d\x94\u20af})$ is abelian, because: $$\mathrm{exp}\beta \x81\u2018({X}_{1})\beta \x81\u2019\mathrm{exp}\beta \x81\u2018({X}_{2})=\mathrm{exp}\beta \x81\u2018({X}_{1}+{X}_{2})=\mathrm{exp}\beta \x81\u2018({X}_{2}+{X}_{1})=\mathrm{exp}\beta \x81\u2018({X}_{2})\beta \x81\u2019\mathrm{exp}\beta \x81\u2018({X}_{1}).$$
To prove claim 2, we showed earlier (as an exercise) that the topological closure of a subgroup is a subgroup, so we just need to show that $\stackrel{{\rm B}\u2015}{\mathrm{exp}\beta \x81\u2018(\mathrm{\pi \x9d\x94\u20af})}$ is compact and abelian. The fact that it's compact follows from the fact that it's closed (by construction) and contained in a compact set (the compact group $G$ ), and it's a general fact from pointset topology that a closed subset of a compact set is compact. (This is clear for us because "compact" means "closed and bounded", and a subset of a bounded set is bounded).
So the key thing to check is that $\stackrel{{\rm B}\u2015}{\mathrm{exp}\beta \x81\u2018(\mathrm{\pi \x9d\x94\u20af})}$ is abelian. This follows from the next lemma below.
Note that it's not always true that exp of a Lie subalgebra gives a subgroup of $G$ .
If $H\beta \x8a\x82G$ is an abelian subgroup then its topological closure $\stackrel{{\rm B}\u2015}{H}$ is abelian.
$\stackrel{{\rm B}\u2015}{H}$ consists of limit points of sequences ${h}_{k}$ in $H$ which converge in $G$ . Suppose I have sequences ${g}_{k}\beta \x86\x92g$ and ${h}_{k}\beta \x86\x92h$ with ${g}_{k},{h}_{k}\beta \x88\x88H$ , $g,h\beta \x88\x88\stackrel{{\rm B}\u2015}{H}$ . Since $H$ is abelian, we have $${g}_{k}\beta \x81\u2019{h}_{k}={h}_{k}\beta \x81\u2019{g}_{k}\beta \x81\u2019\text{\Beta for all\Beta}\beta \x81\u2019k.$$ This implies that $$lim\beta \x81\u2018{g}_{k}\beta \x81\u2019{h}_{k}=lim\beta \x81\u2018{h}_{k}\beta \x81\u2019{g}_{k},$$ so it's enough to show that $$lim\beta \x81\u2018({g}_{k}\beta \x81\u2019{h}_{k})=g\beta \x81\u2019h\beta \x81\u2019\text{\Beta and\Beta}\beta \x81\u2019lim\beta \x81\u2018({h}_{k}\beta \x81\u2019{g}_{k})=h\beta \x81\u2019g.$$
This follows from the fact that matrix multiplication is continuous. Let $m:G\beta \x81\u2019L\beta \x81\u2019(n,\mathrm{\pi \x9d\x90\x91})\Gamma \x97G\beta \x81\u2019L\beta \x81\u2019(n,\mathrm{\pi \x9d\x90\x91})\beta \x86\x92G\beta \x81\u2019L\beta \x81\u2019(n,\mathrm{\pi \x9d\x90\x91})$ be matrix multiplication, i.e. $m\beta \x81\u2019(X,Y)=X\beta \x81\u2019Y$ . This is continuous: it's actually a polynomial in the matrix entries. Therefore $$lim\beta \x81\u2018m\beta \x81\u2019({g}_{k},{h}_{k})=m\beta \x81\u2019(lim\beta \x81\u2018{g}_{k},lim\beta \x81\u2018{h}_{k})$$ (you can "bring limits inside" continuous functions). This means exactly that $${g}_{k}\beta \x81\u2019{h}_{k}\beta \x86\x92g\beta \x81\u2019h,$$ as required.
The proposition now follows by taking the pathcomponent of the identity inside $\stackrel{{\rm B}\u2015}{\mathrm{exp}\beta \x81\u2018(\mathrm{\pi \x9d\x94\u20af})}$ .
Example
The following example should illustrate why I went to the bother of taking the topological closure and what it buys for us.
Let $G$ be the 2dimensional torus $U\beta \x81\u2019(1)\Gamma \x97U\beta \x81\u2019(1)$ . We can draw it as a square with its opposite sides identified (the square lives in the $({\mathrm{\Xi \u0388}}_{1},{\mathrm{\Xi \u0388}}_{2})$ plane, and the sideidentifications correspond to when ${\mathrm{\Xi \u0388}}_{1}$ or ${\mathrm{\Xi \u0388}}_{2}$ cross over a multiple of $2\beta \x81\u2019\mathrm{{\rm O}\x80}$ and restart from zero.
On the square, a 1parameter subgroup of $G$ will look like a bunch of straight line segments with the property that when one of the segments "leaves the square", another one reenters the square at the opposite point (with the same slope).
Remember that a 1parameter subgroup is precisely exp of a line in the Lie algebra. Pick a 1parameter subgroup corresponding to a line of irrational slope. This will wrap around infinitely often in the group $G$ : in fact, it is dense in $G$ (comes arbitrarily close to any given point). The topological closure of this subgroup is then the whole of $G$ .