To Prove the statement
non-trivial group has at least two distinct representations with degree 1.
I would like to use induction. Also I have the relation $$|G|=\sum_{k=1}^{n} d_k^2$$, where n stands for degree of irreducible representation and $d_k$ stands for degree of $k$th representation.
Since G is not trivial, so suppose $|G|=2$, then clearly we have $2=1^2+1^2$ is the only possibility.
Suppose ture for $|G| = m$, then $$|G|=\sum_{k=1}^{n} d_k^2 = 1+1+ \sum_{k=3}^{n} d_k^2$$
Now if $|G|=m+1$, then $$|G|=1+1+1+\sum_{k=3}^{n} d_k^2$$, where the 3=1+1+1 is the only possibility.
Therefore, I think I finished my proof by induction. However, I am not confident with this proof...I am thinking if I misunderstood something or missed some crucial parts?
Could someone help me to check this proof? Or provide me some hints to prove the statement in different way.
Thanks in advance!
If water is wet then snow is hot. If grass is green then pandas are red. If this is a math site then Siddhārtha Gautama is the President of the United States. In the implication "If X then Y," you can't just set X and Y to be any two claims you want! The claim Y has to logically follow from claim X for some reason.
In your "If X then Y" implication, you set X to be the claim "groups of order $m$ have two distinct one-dimensional representations" and Y to be the same claim but for $m+1$. But why in the world do you think that a group of order $m+1$ would have two distinct one-dimensional representations just because groups of order $m$ do? You can't just make stuff up.
Induction wouldn't really be the way to go, because generally groups of order $m$ and $m+1$ don't have anything to do with each other any more than the prime factorizations of $m$ and $m+1$.
In fact, the claim that you're trying to prove
is false.
If $G$ is any nonabelian finite simple group, then a nontrivial representation $G\to\mathrm{GL}_1(\mathbb{C})$ must have a kernel smaller than $G$, but since $G$ has no proper nontrivial normal subgroups and the kernel is normal, the kernel must be $1$, so by the first isomorphism theorem $G$ is isomorphic to a subgroup of $\mathrm{GL}_1(\mathbb{C})$, but $\mathrm{GL}_1(\mathbb{C})$ is abelian and subgroups of abelian groups are abelian hence $G$ is abelian, a contradiction.
The question in your title,
is a different question. The answer is yes. This MathOverflow thread discusses this exact question. The proof seems to depend on the CFSG (classification of finite simple groups), so is pretty hard.